# PROGRESSION

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```					              ARITHMETIC PROGRESSION                                            GEOMETRIC PROGRESSION

Let us study the following sequence :                            Let us study the following sequence :
a, ar, ar2, ar3 …………..
a, (a + d), (a + 2d), (a + 3d) …….
In the above sequence,
In the above sequence,
t1 = a
t1 = a                                                           t2 = ar
t2 = a + d                                                       t3 = ar2
t3 = a + 2d                                                      t4 = ar3
t4 = a + 3d                                                      ---------
--------                                                         ---------
--------
Let us check the ratio between successive terms. i.e.,
Now let us check the difference between successive terms.
t 2 ar       t 3 ar 2     t 4 ar 3
i.e. t2 – t1 = a + d – a = d                                          r,             r,  2  r
t1 a        t 2 ar        t 3 ar
t3 – t2 = (a + 2d) – (a + d) = d                             such a progression is called Geometric Progression
t4 – t3 = (a + 3d) – (a + 2d) = d
Definition :
Such a progression is called Arithmetic Progression.
Geometric Progression is a sequence in which the ratio between
Definition : Arithmetic Progression is a sequence in which the   any two successive terms is constant. This constant ratio is
difference between any two successive terms is constant. This    called common ratio.
constant difference is called common difference.
General representation
General representation
Arithmetic progression is generally represented as               Geometric progression is represented as
a, (a + d), (a + 2d), (a + 3d) …….                               a, ar, ar2, ar3, ar4…
Where ‘a’ is the first term and ‘d’ is the common difference.    Where ‘a’ is the first term and ‘r’ is the common ratio.
General term of an A. P.                                                           General term of a G. P.

In the sequence a, a+d, a+2d, a + 3d ……….                           In the sequence a, ar, ar2, ar3, ar4 ……….
t1 = a     = a + (1 – 1) d                                          t1 = a = ar1-1
t2 = a + d = a + (2 – 1) d
t2 = ar = ar2-1
t3 = a + 2d = a + (3 – 1) d
t4 = a + 3d = a + (4 – 1) d                                         t3 = ar2 = ar3 -1
-------------------                                                 -------------
-------------------                                                 tn = arn – 1
tn = a + (n – 1) d                                                  Hence the general term (nth term) of a G. P. = tn = arn-1
Hence the general term (nth term) of an A. P. = tn = a + (n – 1)d
Sum to n terms of a Geometric Progression
Sum to n terms of an Arithmetic Progression                         Let ‘a’ be the first term and ‘r’ be the common ratio. Then the
Let ‘a’ be the first term, ‘d’ the common difference and , the      sum of the ‘n’ terms in a G. P. is ………..
last term of an A.P. Then the sum of ‘n’ terms is given by          Sn = a + ar + ar2 + ……… + arn-1 ……… (1)
Sn = a + (a + d) + (a + 2d) + (a + 3d) + …………. + ( – 2d)
Now multiplying by ‘r’ on both the sides, we get
+( –d )+                                    ……. (1)
Let us reverse the sequence ……….                                    Snr = ar + ar2 + …… + arn – 1 + arn ……… (2)
Sn = + ( - d) + ( - 2d) + ( - 3d) + ………… + (a + 2d)                 Subtracting (2) from (1)
+ (a + d) + a                          …………. (2)
Sn – Snr = (a + ar + ar2 + ……… + arn-1) –
Adding (1) and (2)                                                  (ar + ar2 + ……. + arn-1 + arn) = a – arn
2Sn = (a + ) + (a + d +    - d) + (a + 2d +   - 2d) + ……… +
a (1  r n )
 Sn[1 – r] = a[1 – r ]  Sn =
n
……. ( - 2d + a + 2d) + ( - d + a + d) + ( + a)
1 r
 2Sn= ( + a) + ( + a) + ( + a) + ……… + ( + a) + ( + a)                                                        a (1  r n )
 the sum to n terms of a G. P. = Sn =                  where r  1
 2Sn = n (a + )                                                                                                  1 r
n           n                                                                       Infinite geometric progression
Sn =
2
 a    2 [ First term + Last term]
Let be the nth term of the sequence                                    If | r | < 1, then the sum of infinite terms will not exceed a
a
 = tn = a (n – 1) d                                                   particular number, which is
1 r
n
 Sn = [a + a + (n – 1)d]                                                                                                      a
2                                                                 S = 1 + r + r2 + ………… + rn + …….  =                    ( | r | < 1)
1 r
n
 the sum to n terms of an A. P. = Sn =     [2a + (n – 1)d]
2
Properties
Properties
1. If all the terms of a G. P. are multiplied or divided by the
1. If each of the terms of an A. P. are increased or decreased by          same quantity, the resulting sequence will be in G. P. with
a constant quantity, the resulting sequence will be in A. P.,          the same common ratio as before.
with the same common difference as before.                         Ex. If 1, 2, 22, 23, …….. are in G. P., then let us multiply each
Ex: If 1, 3, 5, 7, 9 …….. are in A. P. then let us add 2 to each           term with ‘3’. It becomes 3, 6, 12, 24, ……. The resulting
term. It becomes 3, 5, 7, 9, 11, …….… The resulting
sequence will also be in G. P. with a common ratio 2.
sequence will also be in A. P. with a common difference ‘2’.
Now let us subtract ‘2’ from each term. It becomes -1, 1, 3,           Now let us divide each term by ‘3’. It becomes
5, 7, ….. The resulting sequence will also be in A. P., with a          1 2 2 2 23
, , , , ……. The resulting sequence will also be in
common difference 2.                                                    3 3 3 3
2. If each of the terms of an A. P. are multiplied or divided by           G. P., with a common ratio ‘2’.
a constant quantity, the resulting sequence will be in A. P.,
2. The reciprocals of the terms of a G. P. are also in G. P.
with a common difference equal to that of the given
sequence, multiplied or divided by the corresponding                   If each term of a G. P. is raised to the same power, the
constant quantity.                                                     resulting terms will also form a G. P.
Ex: If 5, 8, 11, 14, 17, ….. are in A. P., then let us multiply each                                              1 1 1
Ex: If 1, 3, 32, …….. are in G. P., then , , 2 , …… will be
term with 2. It becomes 10, 16, 22, 28, 34, …… The                                                            1 3 3
resulting sequence will also be in A. P., with a common                                                1
difference ‘6’.                                                        in G. P. with a common ratio .
3
Now let us raise the power to 2. It becomes (1)2, (3)2,  32  ...
2
5 8 11
Now let us divide each term with 2. It becomes , , ,…
2 2 2
i.e., 1, 32, 34, …… forms a G. P. with a common ratio (3)2.
The resulting sequence will also be in A. P., with a common
3. If a1, a2, a3, ….. an-2, an-1, an is a geometric progression, then
3
difference .                                                             a1an = a2an-1 = a3an-2 = …..
2
3. If a1, a2, a3, ……. an-2, an-1, an is an arithmetic progression,          i.e., product of two terms, which are equidistant from both
then a1 + an = a2 + an-1 = a3 + an-2 = ………                              the end terms, is equal.
i.e. the sum of two terms, which are equidistant from both          Ex: If 1, 3, 32, ………38, 39, 310 are in G. P., then 1  310 = 3  39
the end terms, is equal.                                                = 32  38 = ….
Ex. If 10, 20, 30, …… 80, 90, 100 are in A. P. , then                   4. If a1, a2, a3, a4, ….. and b1, b2, b3, b4, ……are two geometric

10 + 100 = 20 + 90 = 30 + 80 = ………                                      progressions, then the sequence, a1b1, a2b2, a3b3, a4b4,…..

4. If a1, a2, a3, a4 …….. is an A. P. and b1, b2, b3 ……. is                                                a1 a 2 a 3 a 4   a
will be in G. P. and also        , , , ,........ n will be in
b1 b2 b3 b4      bn
another A. P., then a1+b1, a2+b2, a3+b3, ..… is also in A. P.
G. P. We can conclude that multiplication or division of the
And also a1 – b1, a2 – b2, a3 – b3 ….. is in A. P.
respective terms of two G. P’s will also be in G. P.
We can conclude that the sum or difference of the respective
Ex: If 1, 2, 22, 23, ….. and 1, 3, 32, 33, ….. are two geometric
terms of two A. P.’s is also in A. P.
progressions, then 1  1, 2  3, 22  32, 23  33, …. are in G.P.
Ex: If 1, 3, 5, 7, …. are in A. P. and 5, 8, 11, 14, ….. are in A. P.
i.e., 1, 6, 36, 216 … are in G. P. with a common ratio 6.
then, 1 + 5, 3 + 8, 5 + 11, 7 + 14, …… are also in A. P. i.e.,
1 2 2 2 23                       1 2 4 8
6, 11, 16, 21, …. are in A. P. with a common difference ‘5’.            and , , 2 , 3 ,…… are in G. P. i.e. , , , , …. are
1 3 3 3                          1 3 9 27
and 1 – 5, 3 – 8, 5 – 11, 7 – 14, …… are also in A. P. i.e.,                                         2
in G. P. with a common ratio
-4, -5, -6, -7, …. are in A. P. with a common difference -1.                                         3
Arithmetic Mean (A. M.)
Geometric Mean (G. M)
If any three quantities are in A. P., then the middle term is said
to be the arithmetic mean of the other two i.e., if ‘A’ is the       If any three quantities are in G. P., then the middle term is said
arithmetic mean between ‘a’ and ‘b’, then a, A, b are in A. P.       to be the geometric mean of the other two. i.e., if ‘G’ is the
ab                                                        geometric mean between ‘a’ and ‘b’, then a, G, b are in G. P.
and A =
2                                                        and G2 = ab or G = ab
In an arithmetic progression, the first and last terms are called    In geometric progression, the first and last terms are called
extremes. The terms present in between are called arithmetic         extremes. The terms present in between are called geometric
means.                                                               means.
Summary of the formulae                                             Summary of the formulae

Formula          Variables in
Formula          Variables in the
the formula
formula
General term (tn)     tn = a + (n -1)d        tn, a, n and d
General term (tn)        tn = a rn-1          tn, a, r and n
n
Sum to ‘n’ terms (Sn) Sn = [2a+ (n -1)d] Sn, n, a and d                                            a ( r n  1)
2                                        Sum to ‘n’ terms (Sn)    Sn =                 Sn, a, r, n
( r  1)
ab           A, a, b
Arithmetic mean (A) A =                                                                               a
2                                      Sum to infinity ( S )   S =                 S , a, r
1 r
Geometric mean (G)       G=      ab           G, a, b
SIMPLE APPLICATIONS
Mode 1

We know the general form of A.P as                                   We know the general form of G. P. as a, ar, ar2, …… arn-1.
a, a+d, ……… a + (n – 1)d.                                            Here also the sequence can be generated by knowing two terms
Here the sequence can be generated by knowing two terms ‘a’          ‘a’ and ‘r’. a and r are also called the parameters of the
and ‘d’. a and d are also called the parameters of the               progression.
progression.                                                         In this model, given the parameter of a G. P., we are expected to
In this model, given the parameters of an A. P., we are expected     find.
to find.                                                             (i) the sequence
(i) the sequence                                                     (ii) the nth term
(ii) the nth term                                                    (iii) expression for sum to n terms
(iii) the expression for sum to n terms
S5) If the first term and the common ratio in a G. P. are 1
S4) If the first term and common difference in an A.P. are
8 and -1 respectively, then find                                     and 3 respectively, find (i) the general term (ii) the
(i) the general term          (ii) the progression (iii) the         progression (iii) the 12th term and (iv) the expression for
th                                                                sum to n terms and hence sum to 10 terms
10 term and (iv) the expression for sum to n terms and
hence sum to 10 terms.                                          Sol: Given the first term = a = 1
Sol: Given the first term = a = 8 and the common difference = d          The common ratio = r = 3
= -1                                                            (i) The nth term of a G.P. = tn = arn-1
(i) The nth term of an A. P. = tn = a+(n -1) d                           Substituting a and r in tn

 3               3
substituting a and d in tn                                                              n 1              n 1
 t n  1×                 
 tn = 8 + (n-1)(-1) = 8 – n + 1  9 – n
 3
n 1
(ii) By substituting n = 1, 2, 3, … in the general term tn = 9 – n   (ii) By substituting n = 1, 2, 3, … in the general term tn=
we can generate the arithmetic progression.
we can generate the geometric progression
 3
11
n = 1  t1                           =1
n = 1  t1 = 9 – 1 = 8
=  3
21
n = 2  t2 = 9 – 2 = 7                                             n = 2  t2                                 3
n = 3  t3 = 9 – 3 = 6
  3                     3
31               2

n = 4  t4 = 9 – 4 = 5                                             n = 3  t3                            

  3                     3
41
Hence the progression is 8, 7, 6, 5 ………..                          n = 4  t4
3
 3 3
(iii) The 10th term of the A. P. can be calculated by substituting
n = 10 in the nth term  t10 = 9 – n = 9 – 10 = -1                 Hence the progression is 1,                      3 , 3, 3 3 , ………..
n
(iv) The sum to ‘n’ terms of an A. P. = Sn =  2a  ( n  1)d       (iii) The 12th term of the G. P. can be calculated by substituting
2                        n = 12 in the nth term of the G. P.

 3                    3
Substituting ‘a’ and ‘d’ in Sn, we get,                                               n 1                    11
 t12                                          243 3
n                         n
Sn =  2×8  ( n  1)( 1)  16  ( n  1)
2                         2
(iv) The sum to ‘n’ terms of a G.P. = Sn =

a rn  1    
n                n
= 16  n  1)  17  n                                                                                                      r 1
2                2                                         Substituting ‘a’ and ‘r’ in Sn, we get,

 3   1  3   1
We get the sum to 10 terms by substituting n = 10 in the
1
n                                 n
above expression                                                                    
Sn               
10
 S10  17  10 = 35
2
 3  1      3  1
We get the sum to 10 terms by substituting n = 10 in the
In this model, given an Arithmetic Progression, we are
above expression
expected to find
 
10
1
          
(i) the parameters (a and d)                                                          3         242
(ii) the general term           (iii) the sum to ‘n’ terms           S10                              121 3  1

3 1        3 1
S6) Find the (i) first term and common difference                        In this model, given a Geometric Progression, we are
(ii) general term (iii) sum to ‘n’ terms and hence sum to            expected to find (i) the parameters (a and d)
10 terms of the A. P., x – 2b, x + 2b, x + 6b, …….                   (ii) the general term (iii) the sum to ‘n’ terms
Sol: Given the A. P., x – 2b, x + 2b, x + 6b, ……….                   S7) Find the (i) first term (ii) common ratio
(i) The first term = t1 = a = x – 2b                                      (iii) general term       (iv) sum to ‘n’ terms and hence the
The common difference = d = t2 – t1 = x + 2b – (x – 2b) = 4b                                         x 1 y
Sum to 10 terms of the G. P. , , 3 …….
(ii) The general term of the A. P., tn = a +(n – 1)d ,                                                     y x x
Substituting ‘a’ and ‘d’ in tn                                                     x 1 y
Sol: Given G. P. , , 3 …………
 tn = (x – 2b) + (n -1)4b = x – 6b + 4nb                                          y x x
n                                                     x
(iii) The sum to ‘n’ terms of an A. P. = Sn = [2a + (n – 1)d]        (i) The first term = t1 = a =
2                                                     y
Substituting a and ‘d’ in Sn                                                                     1
t2           y
n                                                          The common ratio =         x  2
 Sn   2( x  2b )  ( n  1)4b                                                        t1 x         x
2                                                                                         y
= n [ x – 2b + (n – 1)2b] = n [ x + 2nb – 4b] … (1)            (ii) The general term of the G. P. = tn = arn-1
Substituting n = 10 in (1) we get,                                                                                 n 1
x y                   yn2
S10 = 10 x + 160 b                                              Substituting ‘a’ and ‘r’ in tn.  t n   2           
y x                   x 2n 3
S8) Find the sum of the first ‘n’ natural numbers and then find
the sum of first 100 natural numbers.                                (iii) The sum to ‘n’ terms of a G.P. = Sn =

a rn  1         
r 1
Sol:The natural numbers are 1, 2, 3, 4, ………….
x  y         
n
These are in A. P. with first term = a = 1
  2   1
and common difference = d = 1                                                                                   y  x         
n                             Substituting ‘a’ and ‘r’ in Sn.  Sn                  …. (1)
The sum to ‘n’ terms of an A. P. = Sn = [2a + (n – 1)d]                                                             y 
2                                                                        x2   1
 
Substituting a = d = 1, the sum of first ‘n’ natural numbers
 x   y       
10
n                     n ( n  1)
= Sn   2  1  ( n  1)1              …. (1)                                                                          2   1
 y  x          
2                          2                                                                                           
Substituting n = 100 in (1), we get,                                     Substituting n=10 in (1), we get,  S10 
 y       
100(100  1)                                                                                                 x 2  1
S100                   = 5050                                                                                               
2
3 3       3
S9) Find the sum to infinity of the G. P. - , ,          , ……
4 16     64
Sol: From the given geometric sequence, the first term =
3
a= 
4
3
t            3 4          1
Common Ratio = r = 2  16                    
t1  3      16 3         4
4
a
The sum to infinity of a G. P = S =
1 r
By substituting the values of ‘a’ and ‘r’ in the formula, we
3         3       3
get, S         4         4       4   3 4   3
 1       1
1      5         4 5        5
1   
 4          4      4

Model - III
In this model the expression for sum to ‘n’ terms of the progression is given and we have to find the general terms and
generate the progression

S10) If the sum of ‘n’ terms of an A. P. is 2n + 3n2, generate         S11) If the sum of n terms of a G. P. is 5(2n – 1), generate the
the progression and find the nth term.                                 progression and find the general term.
Sol: Given Sn = 2n + 3n2                                               Sol: Given Sn = 5(2n – 1)
Substitute     n = 1  S1 = 2(1) + 3(1)2 = 5                           Substitute     n = 1  S1= 5(2t – 1) = 5
n = 2  S2 = 2(2) + 3(2)2 = 16                                         n = 2  S2 = 5(22 – 1) = 15
n = 3  S3 = 2(3) + 3(3)2 = 33                                         n = 3  S3 = 5(23 – 1) = 35
n = 4  S4 = 2(4) + 3(4)2 = 56                                         n = 4  S4 = 5(24 – 1) = 75
S1 = Sum to 1st term is nothing but the first term itself (t1)        S1 = sum to 1st term is nothing but the first term itself(t1)
S2 = Sum of first two terms t1 and t2                                 S2 = Sum to first two terms t1 and t2
Similarly       S3 = sum of first three terms t1, t2 and t3.          Similarly S3 = sum of first three terms t1, t2 and t3
S4 = sum of first four terms t1, t2 t3 and t4.                  S4 = sum of first four terms t1, t2, t3 and t4
 S1 = t1 = 5                                                S 1 = t1 = 5
S2 = t1+ t2 = 16                                              S2 = t1 + t2 = 15
S3 = t1 + t2 + t3 = 33                                        S3 = t1 + t2 + t3 = 35
S4 = t1 + t2 + t3 + t4 = 56                                   S4 = t1 + t2 + t3 + t4 = 75
 S2 – S1 = (t1+t2) – t1 = 16 – 5 = 11 = t2
S3 – S2 = (t1+t2+t3) – (t1+ t2) = 33 – 16 = 17 = t3                    S2 – S1 = (t1 + t2) – t1 = 15 – 5 = 10 = t2
Hence the sequence is 5, 11, 17, …..                                         S3 – S2 = (t1 + t2+ t3) – (t1 + t2) = 35 – 15 = 20 = t3
Where a = 5 and d = 6                                                    Hence the sequence is 5, 10, 20, ……………
The general term = tn = a + (n – 1)d                                     Where a = 5 and r = 2
 tn = 5 + (n – 1)6 = 6n – 1                                          The general term = tn = arn-1  tn = 5 × (2)n-1

A6) If the first term and common ratio of a G. P. are 1 and 1.2
A5) If the first term and common difference of an A. P. are 1
respectively, then find (i) the general term (ii) the sequence
and 3 respectively, then find (i) the general term
(iii) the 12th term. (iv) the expression for sum to n terms
(ii) the sequence (iii) the 12th term (iv) the expression for
and hence the sum to ’10 terms.
sum to n terms and hence the sum to 50 terms
A7) The A. P. is given as 10, 17, 24, …….. then find                                                 3 3 3
A8) If the G. P. is given as , , , ……… then find
(i)The parameters (a and d) (ii) the general term and hence                                     2 4 8
the 8th term (iii) the expression for sum to ‘n’ terms and         (i) The parameters (a and r) (ii) the general term and hence
hence sum to 10 terms.                                                  the 10th term. (iii) the expression for sum to n terms
A9) If the sum to ‘n’ terms of a sequence is 3n2 + 5n, prove that              and hence sum to 5 terms
it is in A.P., and find its nth term.                                                                                        7
A10) If the sum of ‘n’ terms of a G. P. is given by (3n – 1),
2
then prove that it is in G. P. and find its nth term.
Problems based on solving of equations
(i) Equations with one variable
In this model using the condition given in the problem an equation in a single variable has to be generated.
Solving of this equation gives us the required solution

S12) Which term in the A. P. 5, 2, -1 ……… is – 22 ? Find                   S13)       Which         term              of     the     G.      P.
the sum of these terms.                                                    2 , 6 , 3 2 , 3 6 ..................is 243 2 ? Find the sum
of these terms.
Sol: (i) In the given A.P., the first term = a = 5 and the common
Sol: In the given G. P., the first term = a = 2              and the
difference = d = 2 – 5 = -3                                                                       6
common ratio = r =             3
Let the nth term of the A. P. be – 22                                                             2
 t n  a  ( n  d )  22                                           Let the nth term of the G. P. be 243 2

  
n 1
 5  ( n  1)( 3)  22                                                t n  a ( r )n 1  243 2                    2     3           243 2
 5  ( 3n  3)  22  8  3n  22
 3                                3
n 1                               10
 243  35                       n  1  10  n  11
30
 3n  22  8  n          10
3                                               the 11th term of the G. P. is 243 2
 the 10th term of the A. P. is -22                                 (ii) Substituting a = 2 , r = 3 and n = 11 in the formula
a ( r n  1)
(ii)   Substituting a = 5, d = -3 and n = 10 in the formula                       Sn =
r 1
 
n
2  3  1
11
Sn =     [2a  ( n  1)d] , we get the sum of the 10 terms                                           
2
 S11                    
10                                                                              3 1
       S10 =
2
[2 × 5 + (10 – 1) (–3)] = – 85
=

2 35 3  1      

2 243 3  1       
3 1                        3 1
S14) How many odd integers beginning with 15 must be              S15) How many terms of the geometric sequence 3 , 3, 3 3
taken for their sum to be equal to 975 ?                           , … must be taken to make the sum 39 + 13 3 ?
Sol: The odd integers beginning with 15 are as follows :          Sol: (i) The sum of the terms is 3 + 3 + 3 3 + ……… which
15, 17, 19, ………                                                    forms a G. P. with the first term = a = 3
This forms an A. P. with first term, a = 15 and the common                                        t      3
and the common ratio = r = 2             3
t1      3
difference, d = 17 – 15 = 2
Let n terms of G.P. be taken to make the sum 39+13 3
Let ‘n’ terms of the A. P. be taken to make the sum 975.
 15 + 17 + 19 + ……. N terms = 975                               Sn        3  3  3 3  ......  39  13 3
n                                                             a ( r n  1)
 Sn  (2a + (n-1)d) = 975                                      Sn                     39  13 3
2                                                                r 1
Substituting the value of ‘a’ and ‘d’ in Sn                      Substituting the values of ‘a’ and ‘r’ in Sn
n
 [2×15+(n-1)2] = 975
 
3  3  1
n

2                                                                
               
  39  13 3
 15n + (n -1)n = 975                                          
3 1
 15n + n2 – n = 975
 
 3  3  1  (39  13 3 )( 3  1)
n
               
 n2 + 14n – 975 = 0                                                                

 
 3  3  1  39 3  39  39  13 3
n
 n2 + 39n – 25n – 975 = 0                                                          
               
 n(n+39) – 25(n+39) = 0
  3   3  26 3   3 
n 1                       n 1
 27 3
 (n – 25) (n + 39) = 0
  3   3 . 3   3                        3
n 1       6             n 1                7
 n = 25 or n = – 39                                                                                      
But n = – 39 is rejected since number of terms cannot be          n 1  7  n  6
negative.                                                         the number of terms required to make the sum
 number of odd integers beginning with 15 to make the
39+13 3 = 6
sum equal to 975 = 25
S16) Find the value of ‘K’ if 2K+7, 6K– 2, 8K– 4 are in A. P.        S17) Find the value(s) of ‘P’ if the sum to infinity of the G. P.
Sol: Given that 2K +7, 6K-2 and 8K-4 are in A. P.                              1               25
P, 1, ,.......... is
P               4
The difference between successive terms in an A. P. is                                                   1               25
same                                                            Sol: Given that the sum of the terms P, 1, ,........ is
P               4
 t 2  t1  t 3  t 2                                                                                              1
The first term= a=P, the common ratio = r = and S  25
 (6K  2)  ( 2K  7 )  (8K  4)  (6K  2)                                                                       P       4
a
 4K  9  2K  2                                                                                     S 
The sum to infinity of a G. P. =      1 r
7
 2K  7  K                                                        Substituting ‘a’, ‘r’ and S in the above expression
2
Substituting the value of ‘K’ in 2K + 7, 6K – 2, 8K – 4,                  25    P     25    P2
                       4P 2  25( P  1)
7        7       7                                             4       1
1 P   4    P 1
we get 2× +7, 6× -2, 8× -4 i.e., 14, 19, 24
2        2        2
 the sequence is 14, 19, 24 ………..                                     4P 2  25P  25  0  4P 2  20P  5P  25  0
_________________________________________________
 4P( P  5)  5( P  5)  0  ( 4P  5)( P  5)  0  P  5 or 5
A11) Which term of the A. P. 100, 84, 68, ……. is – 108 ?                                              5
4
Find the sum of these terms.                                           the value of ‘P’ are 4 and 5
A13) The first term of the arithmetic progression is unity and the
common difference is 4, is the number 10091 a term of that      A12) Which term of the G. P. 2, 2 2 , 4 ……. is 64 ? Find the
progression ?                                                        sum of these terms.
A14) The first term of a geometric progression is 128 and the
A15) How many terms of the progression 21, 18, 15, …… must                                 1                 1
be taken for their sum to be zero ? Which term of the                common ratio is 2 . Is the number 256 a term of the G.P.
progression is zero ?                                           A16) How many terms of the geometric progression 3,32,33,…
A17) Determine ‘k’ so that k+2, 4k-6 and 3k-2 are the three              are needed to give the sum 120? Find last term of the
consecutive terms of an A. P.                                       sequence.
2      7
A18) Find the values of ‘x’ so that     , x , are three consecutive
7      2
terms of a G. P.
(ii) Equations with two variables
In this model, using the conditions given in the problem, two equations have to be generated. The two equations are in terms
of two unknown variables. Solve the two equations to get the two unknown variables, which gives us the required solution.

S18) The 8th term of an A. P. is 17, and 19th term is 39. Find           S19) The 6th and 13th terms of a G. P. are respectively equal
3
the A. P. and the 25th term.                                             to 24 and     . Find the G. P. and the 20th term.
16
Sol: Given t8 = 17 and t19 = 39                                                                        3
Sol: Given t6 = 24 and t13 =
We know tn = a + (n – 1)d                                                                        16
We know tn = ar  t6 = ar6-1 = ar5 = 24 …...(1)
n-1

 t 8  a  (8  1)d  a  7d  17.........(1)                                                            3
 t13 = ar13-1 = ar12 =            …..(2)
 t19  a  (19  1)d  a  18d  39.....( 2)                                                            16
Equations (1) and (2) are two equations in terms of the                  Equations (1) and (2) are two equations in terms of the
unknown variables ‘a’ and ‘d’. Solving of (1) and (2)                    unknown variables ‘a’ and ‘r’. Solving of (1) and (2) gives
gives the values of ‘a’ and ‘d’.                                         us the values of ‘a’ and ‘r’.
The first step in solving of the equation is eliminating one             The first step in solving of equation is eliminating one of
of the unknowns.                                                         the unknowns.
This first step in solving of the equation is eliminating one            This can be done by dividing equation (2) by equation (1)
of the unknowns.                                                         where the variable ‘a’ is eliminated as follows :
This can be done by subtracting equation (1) from equation                               3
12
( 2)   ar                     3 1
(2) where the variable ‘a’ is eliminated as follows :                          5  16  r 7  
(1)    ar      24            16 24
(2) – (1) = a + 18d – (a + 7d) = 39 – 17  11d = 22 ....(3)                                                        7
3        1    1        1       1
Equation (3) is an equation in single variable ‘d’ and can                           4 3  7  r7     r 
be found out very easily.                                                  16  3  8 2  2  2         2      2
11 d = 22  d = 2                                                        Now how to get ‘a’ ?
Now how to get ‘a’ ?
We can find the value of ‘a’ by substituting the value of ‘d’         We can find the value of ‘a’ by substituting the value of ‘r’
in equation (1).                                                      in the equation (1)
a + 7d = 17  a + 7 × 2 = 17  a = 3                                                        5
1
 The first term of the A. P. = a = 3 and                             ar = 24  a   = 24  a = 24 × 25 = 24× 32 = 768
5

 2
Common difference = d = 2.                                            The first term of the G. P. = a = 768 and common ratio
The A. P. is 3, 5, 7, ……                                                        1
th                                              =r=
Now let us write the 25 term by substituting a = 3, d = 2                       2
and n = 25 in tn = a +(n–1)d  t25 = 3+(25–1)2= 51                    The G. P. is 768, 384, 192, 96 …….
 The 25th term of the A. P. is 51                                                                                                   1
Now let us write the 20th term where a = 768, r =                and
2
S20) The number of terms in an A. P. is even. The sum of                                                                          19
1        3
odd and even numbered terms are 24 and 30,                            n = 20 in tn = ar   n-1
 t20 = ar   20-1
= 768        11
respectively. If the last term exceeds the first term by                                                                   2      2
10.5, the number of terms in the A. P. is ----------                                                           3
 the 20th term of the G. P. is
Sol: Let the number of terms in the A. P. be ‘n’.                                                                  211
The two conditions given in the problem are :
S21) In a G. P. of even number of distinct terms, the sum of
(i) The sum of odd numbered terms in an A. P. is S0 = 24             all terms is 5 times the sum of odd terms. Then the
common ratio of the G. P. is - - - - - - .
and the sum of even numbered terms is Se = 30.
(ii) The last term exceeds first term by 10.5.                  Sol: The sum of all the terms in the G. P. is
We know that the general representation of an A. P. is                           2    3                n   a ( r n  1)
S = a + ar + ar + ar + . . . . . + ar =               . . . . . (1)
a, a + d, a + 2d, a + 3d …….                                                                                  r 1
Sum of odd terms, S0 = t1 + t3 + t5 + . . . . . tn-1
 Sum of odd numbered terms,
= a + ar2 + ar4 + . . . . . . . . . + arn-2 is also in G. P.,
S0 = t1 + t3 + t5 + t7 + ………. = 24
 a + a + 2d + a + 4d + ….. .. = 24 ……. (1)
This is in A.P. with a common difference ‘2d’and there          With the first term, t1 = a, the common ratio = r2 and there
n                                                               n
will be terms in the above A. P. since we have even             will be terms since there are even number of distinct
2                                                               2
number of terms in the A. P.                                    terms.
a  r2      1
n/2
Similarly the sum of even numbered terms,                                
                     n 
  a  r  1

 S0 =
Se = t2 + t4 + t6 + …. = 30                                                 r2  1             r2  1
 a + d + a + 3d + a + 5d + ….. = 30 …… (2)                    Given condition in the problem is Sn = 5S0
This will be in A. P. with a common difference ‘2d’ and         Using (1) and (2) ……
n
there will be terms in the above A. P. since we have
2                                                   a  r n  1       5a  r n  1
even number of terms in the A. P.                                                           
r 1               r2  1
Subtracting (1) from (2), i.e., (a + d + a + 3d + a +5d +…)
– (a + a + 2d + a + 4d + ….) = 30 – 24                           r2 – 1 = 5 (r – 1)
n               n                               r2 – 1 = 5 r – 5
 (d + d + d +… terms)=6  ×d= 6  nd =12… (3)
2               2
The other condition given in the problem is that the last        r2 – 5r + 4 = 0
term exceeds the first term by 10.5. Let nth term be the last    r2 – 4r – r + 4 = 0
term  tn – t1 = 10.5  a + (n – 1)d – a = 10.5
 (n – 1) d = 10.5         ……. (4)                            r(r – 4) – 1 (r – 4) =0
With the two conditions in the problem, we got two               r = 1 or 4
equations (3) and (4). On solving these equations, we can       r  1, as all the terms will become equal if
find the variables ‘d’ & ‘n’. i.e., subtracting (4) from (3),
r=1r=4
we get nd –(n–1)d = 12–10.5  nd– nd+d =1.5  d = 1.5
12                                                        the common ratio of the G. P. is ‘4’.
n =        = 8 [Using (3)]
1 .5
 the number of terms in the A. P. is 8
S22) In an A. P. if mth term is1/n and nth term is 1/m, then      S23) The mth term of a G. P. is’n’ and its nth term is m.
prove that mnth term is 1                                                                        n2
1                        1                 Show that its (2m-n)th term is    .
Sol: Given that mth term is  tm = a + (m – 1) d = … (1)                                              m
n                        n
1                       1                Sol: Given that the mth term is ‘n’
and the nth term is    tn = a + (n – 1)d =   …. (2)
m                       m                         tm = arm-1 = n             …… (1)
and tm= a+ (mn–1)d, to get tmn we should know ‘a’ and ‘d’.
How to get ‘a’ and ‘d’ ?                                             And the nth term is ‘m’
For this let us take equations (1) and (2)                            tn = arn-1 = m             ….. (2)
We can find the first term and the common difference by           and t2m–n = ar2m–n–1
solving the equations (1) and (2)                                 To get t2m-n we should solve the equations (1) and (2) by
i.e. subtracting (2) from (1), we get ,                           dividing (1) by (2). i.e.,
1 1
[a + (m – 1) d] –[ a + (n – 1) d] =                               (1)   ar m1  n            n
n m                                  n 1     r m n    … (3)
mn            1                       ( 2)  ar      m            m
 ( m – 1) d – (n – 1)d =          d
nm          nm                       t 2mn  ar 2mn1  a.r m1 mn
Substituting ‘d’ in the equation (2), we get
1      1         1 1      1     1                   = a . rm-1.rm-n
a + (n – 1)       a                   
mn m             m m mn mn                          But a. rm-1 = n from (1) and
1   1
 the first term and the common difference are,       &                      n
mn mn            rm-n =     from (3)
respectively                                                                 m

Substituting these values in the mnth term i.e.,                                         n   n2
 t2m – n     =n×   
1               1     1    mn   1                                     m m
tmn= a+(mn–1)d=        ( mn  1)                 =1
mn              mn mn mn mn
 the mnth term is ‘1’
1
A19) The first term of an A. P. is -2 and the 10th term is 16.        A20) The first term of a G. P. is -3 and the 6th term is           .
81
Determine the 15th term.
A21) If the 6th term of an A. P. is 12 and 8th term is 22. Find            Determine its 15th term.
(i) the first term (ii) the common difference (iii) the A. P.    A22) If the second term of a G. P. is 18 and the sixth term is
1458 find
A23) Find the first term ‘a1’ and the common difference ‘d’ of
(i) the first term (ii) the common ratio (iii) the G. P.
the A. P. in which a2 + a5 – a3 = 10, and a2 + a9 = 17, where
A24) The sum of first ten terms of a G. P. is S1 and the sum of
a2 = second term, a3 = third term and so on.
the next ten terms is S2. Find the common ratio.

Problems based on Properties

Let us recall all the properties of A. P. ……                          Let us recall all the properties of G. P. …….
1.   If each of the terms of an A.P. are increased or decreased       1.   If all the terms of a G.P. are multiplied or divided by the
by a constant quantity, the resulting sequence is in A. P.            same quantity, the resulting sequence will be in G. P. with
with the same common difference as before.                            the same common ratio as before
2.   If each of the terms of an A. P. are multiplied or divided by    2.   The reciprocals of the terms of a G. P. are also in G. P.
a constant quantity, the resulting sequence will be in A. P.,         If each term of a G. P. is raised to the same power, the
with a common difference equal to that of the given                   resulting terms will also form a G. P.
sequence, multiplies or divided by the corresponding             3.   If a1, a2, a3 … an-2, an-1, an is a geometric progression, then
constant quantity.                                                    a1an = a2an-1 = a3an-2 = ….
3.   If a1, a2, a3, … an-2 , an-1, an is an arithmetic progression         i.e. product of two terms, which are equidistant from both
then a1 + an = a2 + an-1 = a3 + an-2 = …….                            the end terms, is equal.
i.e. the sum of two terms, which are equidistant from both       4.   If a1, a2, a3 … and b1, b2, b3 … are two geometric
the end terms, is equal.                                              progressions, then the sequence,
4.   If a1, a2, a3, a4…..is an A. P. and b1, b2, b3….is another            a1 b1, a2b2, a3b3, a4b4, … is in G. P.
A. P., then a1 + b1, a2 + b2 , a3 + b3, … is in A. P.                            a a a              a
and also 1 , 2 , 3 ,........ n is in G. P.
and also a1 – b1, a2 – b2, a3 – b3 …… is in A. P.                                b1 b2 b3           bn
We can conclude that the sum or different of the respective           We can conclude that multiplication or division of the
terms of two A. P’s are also in A. P.                                 respective terms of two G. P’s are also in G. P.
1      1    1                                                     1 1 1
S24) If        ,      ,        are in A. P., prove that               S25) If  , , are in G. P., prove that a2, b2, c2 are in G. P.
bc ca ab                                                          a b c
a      b      c                                                                 1 1 1
,      ,      are in A. P. (a + b + c  0)                  Sol: Given that , , are in G. P.
bc ca ab                                                                       a b c
1    1       1                                          Let us begin our operation by taking reciprocals of all the
Sol: Given that       ,     ,       are in A. P.
bc ca ab                                               terms of the given G. P. The new sequence is also in G.P.
 a, b, c will be in G. P.
Let us begin the operation by multiplying all the terms of            Now squaring each term and the resulting sequence is …
the given A. P. with (a + b + c). The new sequence is also             a2, b2, c2 will be in G. P.
in A. P.
abc abc abc                                            S27) If x = 1 + a + a2 + …..  (-1 < a < 1)
             ,          ,         arein A.P.
bc         ca        ab                                      And y = 1 + b + b2 + ….  (-1 < b < 1)
a     bc b          ac ab         c                                                                  xy
                ,            ,             will be in A.P.        Prove that 1 + ab + a2b2 + …..  
bc bc ca ac ab ab                                                                                x  y 1

a
 1,
b
 1,
c
 1 will be in A. P.               Sol: Given that 1 + a + a2 + ……  = x
bc       ca        ab                                          This series forms an infinite G. P. with the first term
Let us subtract ‘1’ from each term and the resulting                   = a = 1, common ratio = r = a
sequence is another A. P.                                             Hence the sum to infinity of the series ,
a      1

a
 1  1,
b
 1  1,
c
 11                             S                =x
bc            ca           ab                                          1 r 1 a
1             1

a
,
b
,
c
are in A. P.                                        1 a   a  1           ….. (1)
bc ca ab                                                                          x             x
1   1   1                The sum 1 + b + b2 + ……  = y, forms another series of
S26) If a2, b2, c2 are in A. P., prove that         ,   ,                  infinite G.P. with first term =a =1 and common ratio= r =b.
bc ca ab
Hence the sum to infinity of this series is
are in A. P.
a       1           1                     1
S                 y   1  b  b  1  …. (2)
Sol: Given a2, b2, c2 are in A. P.                                                  1 r 1 b            y                     y
Add (ab + bc + ca) to all the terms and the sequence still                                        2 3                   2
We know that if 1, a, a , a , ….. and 1, b, b , b3, … are
remains in A. P.                                                      two G.P.’s then 1, ab, a2b2, a3b3, …… will be in G. P.
 a 2  ab  bc  ca , b 2  ab  bc  ca , c 2  ab  bc  ac                    according to the property of the G. P.
are in A. P.                                                                       1 + ab + a2b2 + a3b3 + … forms another infinite G. P.
 a(a + b)+ c(a + b), b(b + c)+ a(b + c), c(a + b)+ a (b+c)                       with first term = a = 1 and common ratio = r = ab
are also in A. P.                                                                 Now if – 1 < a < 1 and - 1 < b < 1  - 1 < ab <1
 (a + b) (a + c), (b + c) (a + b), (a + c)(c + b) are in A. P.                    1 + ab + a2 b2 + … = S =
1
Divide all the terms of the A. P. with (a + b)(b + c)(c + a)                                                       1  ab
and the sequence still remains in A. P.                                          But a = 1 –
1
and b  1 
1
[from (1) and (2)]
(a  b )(a  c)         ( b  c)(a  b )          (a  c)(c  b )                          x              y
,                        ,
(a  b )( b  c)(c  a ) (a  b )( b  c)(c  a ) (a  b )( b  c)(c  a )               1              1                    1
                             
are also in A. P.                                                                     1  ab        1  1            ( x  1)( y  1)
1  1   1       1
1      1    1                                                                                 x  y                   xy
         ,     ,   are also in A. P.
bc ca ab                                                                                   xy                     xy               xy
                                               
xy  ( xy  x  y  1)   xy  xy  x  y  1   x  y 1
S28) An A. P. consists of 24 terms a1, a2, …. a24 such that
a1 +a5+a9+a16 + a20 + a24 = 225. Find the sum of the A.P.                    S29) A G. P. consists of 10 terms in which a2a9 + a4a7 + a5a6
= 13824 and a1 + a10 =1539. Find the increasing G. P.
Sol: a1 + a5 + a9 + a16 + a20 + a24 = 225
We know that in an A. P. a1, a2, ……. an containing ‘n’                       Sol: Given that a2a9 + a4a7 + a5a6 = 13824 and a1+ a10 = 1539
terms, the sum of the two terms which are equidistant from                         a10 = (1539 – a1)           …. (1)
both the end terms is equal. i.e.,                                                We know that in G. P. a1, a2, …… an-1, an containing ‘n’
a1 + an = a2 + an-1 = a3+ an-2 ……                                                 terms, the product of two terms which are equidistant from
Applying this to the given A. P. containing 24 terms                              both the end terms is equal i.e.,
We get a1 + a24 = a5 + a20 = a9 + a16 …… (1)                                      a1 an = a2an-1 = …….
Given a1 + a5 + a9 + a16 + a20 + a24 = 225                                        Applying this to the given G. P. containing 10 terms
Rearranging as follows ……..                                                       we get, a1a10 = a2a9 = a3a8 = a4a7 = a5a6    …… (2)
(a1+ a24) + (a5 + a20) + (a9 + a16) = 225 ….. (2)                                 Given a2a9 + a4a7 + a5a6 = 13824             ……. (3)
Substituting (1) in (2), we get                                                   Substituting (2)in(3), we get a1a10 + a1a10 + a1a10 = 13824
(a1 + a24) + (a1 + a24) + (a1 + a24) = 225                                                                          13824
 3(a1a10) = 13824  a1a10 =            = 4608           ….. (4)
 3(a1 + a24) = 225  a1 + a24 = 75                                                                                   3
n                                                    Substituting (1) in (4), we get
We know that Sn =        [First term + Last term]
2                                                    a1 (1539 – a1) = 4608  1539 a1 – a12 = 4608
24                 24                                                 a12 –1539a1+4608 = 0  a12–1536a1– 3a1+ 4608= 0
=
2
a1  a 24  
               2
 75 = 900                                     a1(a1– 1536) –3(a1– 1536) = 0  a1= 3 or a1= 1536
If a1 = 3, then a10 = 1539 – 3 = 1536      [from (1)]
1 1 1                            We get the increasing G. P.
A25) If a, b, c are in A. P., prove that     , ,    are in A. P.
bc ac ab                           a1 = 3 and a10 = 1536
1 1 1                                                           tn = arn-1  t10 = 1536 = 3 × r9        [where a = a1 = 3]
A27) If       , , be in A. P. and (a + b + c  0), prove that
a b c                                                            r = 512 = 2  r = 2
9           9

( b  c) ( c  a ) ( a  b)                                           Hence the increasing G. P. is 3, 6, 12, ….
,        ,         are in A. P.
a        b         c                                                                                      b 1 b
A26) If a, b and c are in G. P. prove that      , ,   are in G. P.
[Hint : Begin the operation by multiplying (a + b + c) with every                                                   a 2 b c2
term]                                                             A28) If a2, b2, c2 are in G. P., prove that bc, ca, ab are in G. P.

Interesting Problems

We can assume the various terms of an A. P. for easy                   We can assume the various terms of a G. P. for easy
simplification as given below :                                      simplification as given below :
(i) If an A. P. consists of three terms, they can be assumed as        (i) If a G. P. consists of three terms, they can be assumed as
a – d, a, a+ d.                                                        a
, a, ar.
(ii) If there are four terms, they can be assumed as a –3d,                  r
a– d, a + d, and a + 3d.                                         (ii) If there are four terms in G. P., they can be assumed as
(iii) If there are five terms, they can be assumed as a – 2d,                a a          3
, , ar, ar .
a – d, a, a + d, a + 2d.                                                3
r r
(iv) If there are six terms, they can be assumed as a – 5d,            (iii) If there are five terms, they can be assumed as   a a ,a, ar,ar2
,
a – 3d, a – d, a + d, a + 3d, a + 5d.                                                                                    r2 r
a a a
(iv) If there are six terms, they can be assumed as        , , , ar,
r5 r3 r
ar3, ar5
S30) The product of three terms of an A. P. is 1620 and their    S31) The product of 3 numbers in G. P. is 1728 and their
sum is 36. Find the A. P.                                        sum is 38. Find the numbers.
Sol: The two conditions given in the problem are                 Sol: The two conditions given in the problem are
(i) The product of three terms which are in A. P. is 1620        (i) The product of three terms which are in G. P. is 1728
(ii) The sum of three terms which are in A. P. is 36             (ii) The sum of three terms which are in G. P. is 38
The problem can be solved easily if we assume three terms        The problem can be solved easily if we assume three terms
as a – d, a and a + d                                            as a , a, ar. The product of the terms is a × a × ar
r                                             r
Product of the terms is (a – d) a (a + d) = 1620
= a3 = 1728  a = 12       ..…. (1)
 a (a2 – d2) = 1620 …. (1)                                      The sum of three terms is
The sum of the terms = a – d + a + a + d = 36
a                       1         
 3a = 36  a = 12 …. (2)                                           + a + ar = 38  a   1  r  = 38 …… (2)
Substituting ‘a’ in (1) to get the value of ‘d’                   r                      r         
Substituting ‘a’ from (1) in (2) to get the value of ‘r’
 12(122 – d2) = 1620
1                1           38 38
 122 – d2 =
1620
= 135                                         a   1  r   38   1  r         
12                                                    r                r            a    12
19
 d2 = 144 – 135 = 9  d =  3                                    r + r + 1 = 6 r  6r + 6r – 19r + 6 = 0
2                     2

The three terms in the A. P. a – d, a, a + d are …..              6r2 – 13r + 6 = 0  6r2 – 9r – 4r + 6 = 0
 3r(2r – 3) - 2(2r – 3) = 0
(i) If d = +3                   (ii) If d = -3                    (2r – 3) (3r – 2) = 0  r = 3 or 2
2   3
a – d = 12 – 3 = 9              a – d = 12 + 3 = 15                                              a
a = 12                          a = 12                           The three terms in the G. P.        , a, ar, are …..
r
a + d = 12 + 3 = 15             a + d = 12 – 3 = 9                             3 a 12                            2 a 12
The three numbers are           The three numbers are             (i) If r =    ,  8             (ii) If r =    ,     18
2 r   3                           3 r  2
9, 12, 15 and form an           15, 12, 9 and form a                                3                                 2
increasing A. P.                decreasing A. P.                  a = 12,   ar = 12× = 18          a = 12, ar    = 12× = 8
2                                 3
The three numbers are            The three numbers are
8, 12, 18 which form an          18, 12, 8 which form a
increasing G. P.                 decreasing G. P.
A29) Find the increasing A. P. the sum of whose first three         A30) The continued product of three numbers in G.P. is 216 and
terms is 27 and the sum of their squares is 275.                    the sum of their products in pairs is 156. Find the numbers.
A31) The sum of four integers in A. P. is 20 and the sum of their   A32) The product of four members is 4096 and the common
squares is 120. Find the numbers.                                   ratio is ‘4’. Find the numbers.

In this type we have to prove the given condition with the          In this type we have to prove the given condition with the
help of terms which are in A. P.                                    help of terms which are in G. P.
S32) If a, b and c are in A. P., prove that                         S33) If a, b and c are in G. P., show that a(b2+c2) = c(a2+b2)
(a + 2b –c) (2b + c – a) (c + a – b) = 4abc
Sol: Given that a, b, c are in A. P.                                Sol : Given that a, b, c are in G. P.
 2b = a + c                         ……. (1)                 Common ratio = 
b c
Using (1) let us prove the condition given in the problem                                  a b
i.e. (a + 2b – c) (2b + c – a) (c + a – b) = 4abc                     b = ac
2
….. (1)
In the above expression                                             Using (1), let us prove the condition given in the problem,
(a + 2b – c) = (a + a + c – c) = 2a           ..….. (2)             i.e., a(b2 + c2) = c(a2 + b2)
(2b + c – a) = (a + c + c – a) = 2c           …… (3)                In the above expression
(c + a – b) = 2b – b = b                      …… (4)                L. H. S. = a(b2+c2) = a(ac + c2) = a2c + ac2
 (2) × (3) × (4) = (a + 2b – c ) (2b + c – a) (c + a – b)            = c(a2 + ac) = c(a2 + b2) = R. H. S.
= 2a × 2c × b = 4abc
A34) If x, y, z are in G. P. , show that
1     1     1    1            (x + 2y + 2z) (x – 2y + 2z) = x2 + 4z2
A33) If a, b, c are in A. P., prove that                
a  b b  a a  c 2b
A36) If (a – b), (b – c) and (c – a) are in G. P., prove that
A35) If (a – b), (b – c), (c – a) are in A. P., prove that b = c         (a + b + c)2 = 3(ab + bc + ca)
Problems on Arithmetic Mean                                          Problems on Geometric Mean
1
S34) Insert ‘4’ arithmetic means between 8 and 33.                   S35) Insert five geometric means between       and 243.
3
Sol: Let the first term be ‘8’ and the last term be 33 and A1, A2,   Sol: Let the first term be 1 and the last term be 243, and
A3, A4 be the four arithmetic means between 8 and 33.                                      3
G1, G2, G3, G4, G5 be the five geometric means between
 8,A1 , A2, A3, A4, 33 are in A.P.
1
and 243.  1 , G1, G2, G3, G4, G5, 243 are in G.P.
As we are inserting ‘4’ arithmetic means in between 8 and            3            3
33, the number of terms will be six                                 As we are inserting ‘5’ geometric means in between
t1 = a = 8                   … (1)                                1
and 243, the number of terms will be seven.
3
t6 = a + 5d = 33             … (2)
 t1 = a = 1        ….. (1)
Substituting (1) in (2)                                                        3
t7 = ar7-1 = ar6 = 243 ….. (2)
 8 + 5d = 33  d = 5
Substituting (1) in (2)
The second, third, fourth and fifth terms of the A. P.
become the four arithmetic means A1, A2, A3, A4.                       ar6 = 1 × r6 = 243  r6 = 243 × 3 = 36  r = 3
3
A1 = t2 = a + d = 8 + 5 = 13                                         G1 = t2 = ar = 1 × 3 = 1, G2 = t3 = ar2 = 1 × 32 = 3
3                           3
A2 = t3 = a + 2d = 8 + 2 × 5 = 18
G3 = t4 = ar3 = 1 × 33 = 9, G4 = t5 = ar4 = 1 × 34 = 27
A3 = t4 = a + 3d = 8 + 3 × 5 = 23                                                    3                           3
5   1    5
A4 = t5 = a + 4d = 8 + 4 × 5 = 28                                    G5 = t6 = ar = × 3 = 81
3
 the four arithmetic means between 8 and 33 are                The 5 geometric means between 1 and 243 are 1,3,9,27,81
13, 18, 23, 28.                                                                                   3
S36) If a, b, c be in A. P. and ‘m’ is the A. M. between ‘a’      S37) If the G. M. between pth and qth terms of a G. P. is
and ‘b’ and ‘n’ is the A. M. between ‘b’ and ‘c’, then            equal to the G. M. between mth and nth terms of the G.
show that ‘b’ is the A. M. between ‘m’ and ‘n’.                   P., show that m + n = p + q
ac
Sol: Given that a, b, c are in A.P.      = b  a+c = 2b ...(1)   Sol: The pth term of the G. P. = tp = arp – 1 and
2
ab                      qth term of the G. P. = tq = arq-1
‘m’ is the A. M. between ‘a’ and ‘b’        = m …. (2)
2                        G.M. between pth & qth term = t p t q  ar p 1  ar q 1 ….(1)
bc                       The mth term of the G. P. = tm = arm-1
‘n’ is the A. M. between ‘b’ and ‘c’        = n ….. (3)
2
mn                                 and nth term of the G. P. = tn = arn-1
A. M. between ‘m’ and ‘n’ =                       …… (4)           G.M. between mth& nth term = t m t n  ar m1  ar n 1 …(2)
2
Substituting the values of ‘m’ & ‘n’ from (2) & (3) in (4)         Given that (1) = (2)       ar p1  ar q1    ar m1  ar n 1
ab bc
mn             
           2         2  a  2b  c  2b  (a  c )              Squaring on both sides
2            2             4            4                    arp-1 × arq-1 = arm-1 × arn-1  rp-1+q-1 = rm-1+n-1
4b          mn
=      b           b                                           As bases are equal, we can equate the powers
4            2
 ‘b’ is the A. M. between ‘m’ and ‘n’                             p+q–2=m+n–2p+q=m+n

A38) Insert 3 geometric means between 4 and 324
A37) Insert 8 arithmetic means between – 5 and 13.
xa        x
xa      xa                            A40) Find the G. M. between       and
A39) The A. M. between         and       is ……….                                                 x      xa
x         x
Arithmetic Progression and Simple Interest                    Geometic Progression and Compound Interest

The total amount after every year with a compound
The simple interest (I) on a principal amount ‘P’ for an        interest on a principal amount ‘P’ for an interest rate ‘r’
PNR
interest rate ‘R’ and for ‘N’ years is given by, I =                                       
N
r 
100        and for ‘N’ years is A = P  1      
Initial amount = P                                                                          100 
PR                          Initial amount = P
The amount after one year = P +         (N = 1)                                                       r 
100                          The amount after one year = P  1       
2PR                                                          100 
The amount after two years = P +          (N = 2)                                                            2
100                                                            r 
3PR                       The amount after two years = P  1    
The amount after three years = P +         (N = 3)                                              100 
100                                                                       3
     r 
Thus the amounts after every year are in A. P. with             The amount after three years = P  1    
PR                                                                       100 
common difference of                                            The amount after every year are in G.P. with common rato
100
     r 
of  1    
S38) A man takes a loan of 6,000/- from a bank. He agrees                100 
to pay back the loan in 12 annual instalments of
Rs.500/- each along with the interest on unpaid            S39) A man saves Rs.1000/- each year upto 10 years. If the
amount. The rate of interest is 12% per annum. Find             rate of interest is 12% per annum and the interest is
the total amount paid by him to the bank.                       compounded annually. Find the amount the man will
receive at the end of the 10th year.
Sol: The loan taken by the man from the bank = 6,000/-
Amount paid in each instalment = 500                       Sol: The principal amount invested for every year = 1000
Total instalments = 12                                          The amount at the end of 10 years for the principal
N
Total amount to be paid by the man at the end of every                                 r 
year = 500 + interest on unpaid amount                          amount, P = A = P  1    
 100 
PNR                        Where P = Principal amount = 1000/-
Simple interest on unpaid amount =                             r = Rate of interest = 12 %
100
Where P = Principal amount or unpaid amount                    N = number of years = 10
N = Number of years = 12 years                         The amount invested in the first year will earn an interest
R = Rate of interest = 12                              for 10 years and the amount in the second year will earn
Let us make a chart to find the amount at the end of every     an interest for 9 years and so on. Let us make a chart to
year :                                                         find the amounts at the end of 10th year.

Year Amount Interest                    Amount at the end of
Year Unpaid Interest on unpaid Instalment Total amount to
invested calculated                   10th year
amount amount I = PNR after every be paid after
100
for years
year (A) every year (A+I)             st                                            10           10
after the end of                                        1      1000        10                  12     112 
1000  1        100  × 10
3
each year                                                                                  100            
1 st
6000   6000  12  1          500    720+500=1220             2nd     1000         9                           9            9
=720                                                                             12   112 
100                                                                             1000  1        100  × 10
3

2 nd
5500   5500  12  1          500    660+500=1160                                                 100            
=660                                     3rd     1000         8             
8
12   112 
8
100                                                                             1000  1        100  × 10
3

3 rd
5000   5000  12  1          500    600+500=1100                                                 100            
=600
100                                               ---    -----       -----                    -----
--- -----          -----         -----         -----
--- -----          -----         -----         -----                 The earnings of the man for every years investment at the
10                 9
th           3  112    112 
 , 10  100  …..
3
end of 10 year are 10 
Thus the payments made by him at the end of every year is                              100            
1220, 1160, 1100, ………                                          The sum of all these earnings will give the total amount
The sum of all the amounts will give the total amount paid     earned by him after 10 years
10                   9                8
by him to the bank in 12 years                                             112      3  112       3  112 
  10   100   10   100   .....
3
= 10 × 
= 1220 + 1160 + 1100 + …                                                   100                           
The above series form an A. P. with first term = a = 1220          The above series forms a geometric progression with first
 100        a ( r n  1)
10
Common difference = d = 1160 – 1220 = – 60                                  3  112 
term a=10          Common ratio r =         Sn=
The sum of the amounts of 12 years can be calculated                           100                      112  ,        r 1
using
 112   100     
10       10

n                        12                                          103        112   1
Sn   2a  ( n  1)d   S12  [2  1220  (12  1)( 60)]         S10 =      100           = 19652.5 Rs. (approx.)

2                         2
= 6[2440 + 11(–60)] = 6[2440 – 660] = 10680                                        100    
 112  1
        
 The total amount paid by him to the bank for a loan of           The total amount saved by the man at the rate of 12% for
Rs. 6000/– is Rs. 10,680                                         Rs.1000 per year for 10 years is 10,000 for which he gets a
return of Rs. 19,652.5

Practical application of Arithmetic and Geometric progressions
S40) A tree in each year grows 4 cm less than it grew in the previous year. If it
grew 1m in the first year. In how many years will it have ceased growing and
what will be its height then?
Sol: Growth or height of the tree in the first year =1m =100cm
Rate of growth = 4cm less than the previous year’s growth Let us make a chart of
height and growth of the tree year wise.
Year Growth of the tree (4cm Final height of the tree at
S41)    A     particle
less to previous year) the end of every year(in cm)
moves 20 metres in
1st              100                    100
nd                                                           the first minute, 19
2           100 – 4 = 96          100 + 96 = 196               mtrs in the second
rd
3            96 – 4 = 92          196 + 92 = 288                          1
”                ”                      ”                     minute, 18 mtrs in
th                                                                      20
n                 0                      Sn                    the third and so on
according to G. P. what distance can it never exceed ?
Sol: Given that the particle moves 20 metres in first minute, let this be t1  t1 = 20 and
the particle moves 19 metres in the second minute, let this be t2  t2 = 19
1                                           1
It moves 18      metres in the third, let this be t3 = 18
20                                          20
1
Hence the distances moved by the particle in every minute are 20, 19, 18           ….
20
Here the terms form a G. P.
Where the first term = a = 20 and The common ratio =

If we observe the above chart the year wise growth is 100, 96, 92, …. 0
This growth is in A. P. with
The first term = a = 100, common difference = d = – 4
The nth term of the sequence = tn = a + (n – 1)d = 0
 100 + (n – 1) ( – 4) = 0
 4(n – 1) = 100  ( n – 1) = 25  n = 26
 after 26 years the tree stops growing
The total height of the tree = Sum of the growth at the end of every year = 100 + 96
+ 92 + ….. 0 (26 terms)
n
We know that the sum of the n terms = Sn= (2a+(n -1)d)
2
26
 S26 =      (2 × 100 + (26 – 1) ( –4) = 13[200 + (25) (–4)]
2
= 13[200 – 100] = 1300 cm
We can also get the height by finding the 26th term of the series of 3rd column. i.e.
100, 196, 288 , ….
S42) A leaf is torn from a paperback novel. The sum of the numbers on the
remaining pages is 3825. If there are 100 pages in the novel, find the page
number of the torn leaf.

Sol: Let the page number of the book from which a leaf is torn = x . Given that the
number of pages in the book = 100
the remaining pages in the book = 100 – x
The sum of these (100 – x) pages =
(x + 1) + (x + 2) + (x + 3) + (x + 4) + …. + 100 = 3825
 the above series form an A. P. where,
the first term = a = (x+1),
1
t             t    18
19              20  361  19
r= 2          or 3 
t1    20     t2      19      19  20    20
1
The total distance travelled = 20 + 19 + 18 + …
20
This becomes the sum of an infinite G. P. as the common ratio is less than 1 
a       20      20
S                         = 400 m
1 r     1
19     1
20    20
Hence the total distance travelled by the particle can never exceed 400 m.

2
S43) A ball is dropped from a height of 48 m and rebounds        of the distance it
3
falls. If it continues to fall and rebound in this way, how far will it travel
before coming to rest ?

2
Sol: Given that a ball is dropped from a height of 48 m and rebounds         of the distance
3
2
it falls i.e. it rebounds to a height 48 ×     = 32 m and again it rebounds to a height
3
2 64
32 ×     ×   m and so on.
3   3
64
The distance forms a G. P. 48, 32,       , . . . . with the first
3

last term = ℓ = 100
n
We know that Sn =       [First term + Last term]
2
 The sum of the (100 – x) pages
(100  x )
= S(100 – x) =            [ x  1  100]  3825
2
 (100 – x) [ 101 + x] = 3825 × 2
 100 × 101 + 100x – 101x – x2 = 7650
 10100–x2–x =7650  x2+ x–10100+7650 = 0
 x2 + x – 2450 = 0  x2+ 50x–49x – 2450 = 0
 x(x+50) – 49(x + 50) = 0
 (x – 49) (x + 50) = 0  x = 49 or – 50
But x –50, as the page number cannot be negative.
 x = 49  the page number of the book from which the leaf is torn = 49

decided to utilize this money as an offering to God, so you start offering one
paisa first day, 2 paisa on second day, 3 paisa on third day and so on for 365
days. Now the question is, can you complete your amount before 365 days ?

Sol: The amount offered on the first day = 1 paisa
The amount offered on the second day = 2 paisa
The amount offered on the third day = 3 paisa and so on
 the above terms form an A. P. 1, 2, 3, 4 ………
Where the first term = a = 1
Common difference = d = 1
Let us calculate the total amount for 365 days
term = a = 48 and the common ratio = r = 2 .
3
This is an infinite G. P. as we do not know after how many bounces it comes to
rest.
 total distance it travelled before coming to rest
64
= 48 + 32 +       + …..
3
a       48       48  3
 S                              = 48 × 3 = 144 m
1 r     1
2      3 2
3
 the total distance travelled by the ball = 144 m.
S45) The number of bacteria in a certain culture doubles every hour. If there were
30 bacteria present in the culture originally, how much bacteria will be
present at the end of the second hour ? 4th        hour ? nth hour?
Sol: The bacteria originally present is 30 and given that it doubles every hour.
After the end of first hour, the bacteria present 30×2=60
After the end of second hour, the bacteria present is 60 × 2 = 120 and so on.
Hence the bacteria present after every hour is 30, 60, 120, …. Which forms a G. P.
with first term a = 30 and
Common ratio = r = 60 = 2
30
 the general term of the G. P. is tn = arn-1 = 30 × (2)n-1         The    bacteria
present in the nth hour = 30 × 2n-1
The bacteria present in the fourth hour
= 30×(2)4-1=30×23=240
1 + 2 + 3 + 4 + …. + 365
n
We know that Sn = [first term + last term]
2
365               366  365
 S365 
2
1  365      2
= 183 × 365
 66795 paise = 667.95 Rs. ( 100 paise=1Rupee)
Whereas the amount given is only Rs.500/-
 the amount can be completed before 365 days.

S46) A child builds a pattern with square building bricks using the sequence of
steps as shown.

The total number of bricks used is 1 + 3 + 5 + …..
(a) How many bricks does the child use on the nth step?
(b) If the child has 60 bricks, how many                            steps   can   be
completed ?

Sol: (a) As the bricks are arranged in the form 1, 3, 5, 7, …. this forms an A. P. and the
bricks in the nth step can be calculated by using the formula, tn = a + (n – 1)d
Where tn = number of bricks in the nth step.
The first term = a = 1
The common difference = d = 2
 tn = 1 + (n – 1)2 = 1 + 2n – 2 = 2n – 1
 the number of bricks in the nth step is 2n – 1

S47) A stone is dropped in a pond, giving rise to wave motion, in the form of
concentric circles. The area of the circle, it is noticed, increases in geometric
progression. Prove that the radius of the circle also increases in geometric
progression.
Sol: Let A1, A2, A3, ….An are the areas of the concentric circles and r1, r2, r3, … rn are
the respective radii of the circles
Given that A1, A2, A3, …. are in G. P. i.e.,
πr12, πr22, πr32, πr42, ……. πrn2 are in G. P.
Now raising the power of all the terms of the G. P. by 1 , still the sequence remains
2
in G. P.
1             1                1                       1

   πr12    
2
, πr2 2    
2
, πr32      2

....... πrn 2      2
are in G .P.
          π r1 , π r2 , π r3 , ........ π rn are in G .P.
Now dividing all the terms of the G. P. by π , still the sequence remains in G. P.
π r1    π r2 π r3            π rn
,       ,      , ......       r1,r2, r3, …rn are in G. P.
π       π      π             π
 the radius of the concentric circles will be in G. P. if the areas of the concentric
circles are in G. P.
S48) The population of a town increases every year by 10% of the population at the
beginning of that year. What is the percentage increase in population at the
beginning of nth year ?
Sol: Let P be the population of the town at beginning of first year. The population
increases every year by 10% of the population at the beginning of that year.
(b) If the child has 60 bricks then it is equal to the sum of the terms in the A. P. 1,
3, 5, 7, …..
n
We know that, Sn = [2a  ( n  1)d]
2
Here Sn = 60, a = 1, d = 2
n
 60 = [2 × 1 + (n – 1)2]  120 = n[2 + 2n – 2]
2
 120 = n[2n]  120 = 2n2  n2 = 60  n = 60
But ‘n’ should be a natural number
 taking the nearest value below 60 to get a perfect square, i.e. 49
 n = 49 = 7, hence the child can build ‘7’ steps with ‘49’ bricks and ‘11’ bricks
will be left over.
A41) A man 50 years old has 8 sons each of them born at equal intervals. The sum of the
ages of the father and the sons is 186 years. What is the age of the eldest son, if the
youngest son is 3 years old ?

A43) 100 railway sleepers are laid at intervals of 1 m apart. If a man can carry only one
sleeper at a time, how many metres must be travelled in collecting all the sleepers
into a pile in the position of the first ?

Let us make a chart of the population at the beginning and end of the year.
Population     at   the Population at the end of the
Year beginning of the year     year
       10 
1               P                   P 1           (1.1)P
      100 

     10 
2             (1.1)P            (1.1)P 1        (1.1) P
2

 100 
    10 
3            (1.1)2P           (1.1)2 P 1       (1.1) P
3

 100 
    10 
4            (1.1)3P           (1.1)3 P 1       (1.1) P
4

 100 
n
The population at the beginning of every year form a G. P. i.e. P, (1.1)P, (1.1)2P,
(1.1)3P, …with the first term = a = P
(1.1)P
common ratio r =           = (1.1)
P
 the population at the beginning of the nth year = the nth term of the above G. P.
 tn = arn-1 = P(1.1)n-1
A42) The vibrations of a spring are damped so that the amplitudes of successive
deflections are in G.P. If the amplitude initially is 12 cm and then 8 cm and so on,
which are in G. P., find the amplitude of 6th deflection.
A44) The distance covered by a certain pendulum bob in succeeding swings form the G.
P., i.e., 16, 12, 9, …. in centimetres respectively. Calculate the total distance
described by the bob before it comes to rest.

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