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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 ab 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) ab 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 11 n = 1 t1 =1 n = 1 t1 = 9 – 1 = 8 = 3 21 n = 2 t2 = 9 – 2 = 7 n = 2 t2 3 n = 3 t3 = 9 – 3 = 6 3 3 31 2 n = 4 t4 = 9 – 4 = 5 n = 3 t3 3 3 41 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 yn2 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=1r=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 m1 n n n m n 1 r m n … (3) mn 1 ( 2) ar m m ( m – 1) d – (n – 1)d = d nm nm t 2mn ar 2mn1 a.r m1 mn 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. bc ca ab a b c a b c 1 1 1 , , are in A. P. (a + b + c 0) Sol: Given that , , are in G. P. bc ca ab a b c 1 1 1 Let us begin our operation by taking reciprocals of all the Sol: Given that , , are in A. P. bc ca ab 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. abc abc abc S27) If x = 1 + a + a2 + ….. (-1 < a < 1) , , arein A.P. bc ca ab And y = 1 + b + b2 + …. (-1 < b < 1) a bc b ac ab c xy , , will be in A.P. Prove that 1 + ab + a2b2 + ….. bc bc ca ac ab ab x y 1 a 1, b 1, c 1 will be in A. P. Sol: Given that 1 + a + a2 + …… = x bc ca ab 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 11 S =x bc ca ab 1 r 1 a 1 1 a , b , c are in A. P. 1 a a 1 ….. (1) bc ca ab 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. bc ca ab 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. bc ca ab 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 ac 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 ab 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) bc The mth term of the G. P. = tm = arm-1 ‘n’ is the A. M. between ‘b’ and ‘c’ = n ….. (3) 2 mn 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 m1 ar n 1 …(2) 2 Substituting the values of ‘m’ & ‘n’ from (2) & (3) in (4) Given that (1) = (2) ar p1 ar q1 ar m1 ar n 1 ab bc mn 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 mn = 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–2p+q=m+n A38) Insert 3 geometric means between 4 and 324 A37) Insert 8 arithmetic means between – 5 and 13. xa x xa xa A40) Find the G. M. between and A39) The A. M. between and is ………. x xa 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 S44) Suppose your father gifted you Rs.500/- on your birthday and you have 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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