# Sequences and Series - PDF

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```					SEQUENCES AND SERIES
See the following sets of numbers, do they look cumbersome?

A A 1,5,9,13,17,21,25,29, '

2   2   2      2       2   2   2   2
B A1 ,2 ,3 , 4 ,5 ,6 ,7 ,8 , '

1 f ff 1 f 1 f 1 f 1 f 1 f
ff 1f ff ff ff ff ff '
ff ff ff ff ff ff ff
f      f f f f f
ff ff ff ff ff ff ff
CA     , ,   ,  ,  ,  ,   ,
10 11 12 13 14 15 16

But look again closely. Do you see a specific order or pattern in each set? Isn’t it easy to
predict any term (say the 20th) in each of the above? We note that in A we add 4 each
time to get the next number, in B we take the square of the next integer and in C we take
the reciprocals of the next integer. Thus the 20th term in A is 77, in B is
1f
ff
ff
20 and in C is ff These are called sequences, where the subsequent terms are decided
2
.
39
by a given relation (which is expressed as a function f(n) ).

SEQUENCE

A sequence is a function f whose domain is the set of all natural numbers (infinite
sequence) or some subset of natural numbers from 1 up to some larger number (finite
sequence). The values f(1), f(2), f(3). . . . are the terms of the sequence.

Let the terms of a sequence be given by the function by f n = n 2 + 1 where n is a natural number A
` a

f 1 = 1 + 1 = 2,                f 2 = 2 + 1 = 5,         f 3 = 3 + 1 = 10,
` a       2                     ` a     2                ` a   2
Here
f 4 = 4 + 1 = 17,          f 5 = 1 + 1 = 26, ……A
` a        2                    ` a    2

The sequence becomes     2, 5, 10, 17, 26, ………………A

1, 5, 9, 13, 17 is a finite sequence with 5 terms.
1, 5, 9, 13, 17, 21,……… is an infinite sequence.

A sequence is represented by its range. f n = an is used to denote the range elements of
` a

the function. a1 , a2 , a3 , ……an , …A are the first term, second term, third term,….nth
term,…. respectively of the sequence.
Example : Give first 3 terms of the sequence an = n 2 + n + 1 .

Solution : Putting n = 1,2,3 in an = n 2 + n + 1, we get
a1 =1 + 1 + 1 = 3,        a2 = 2 + 2 + 1 = 7,    a3 = 3 + 3 + 1 = 13
2                          2                     2

The sequence would be written 3, 7, 13,………
n
fffff
ffff
Example : Give first 3 terms of the sequence an = fffff.ffff
n +1
2
n
fffff
ffff
ffff
fffff
Solution : Putting n = 1,2,3 in an = 2     , we get
n +1
1
ffff 1f              2
fffff 2f             3
fffff 3f
ffff f
ffff              ffff f
ffff
a1 = fffff f, a2 = fffff f, a3 = fffff ff
=                  =            ffff ff
ffff ff
=
1 +1 3              2 +1 5            3 + 1 10
2                  2                  2

1f 2f 3f
f f ffff
The sequence would be written as f, f, ff, AAAAAAA
3 5 10

SERIES
If a1 ,a2 ,a3 ,a4 , ……AA ,an , ……… is a sequence, then the expression of their summation
a1 + a2 + a3 + a4 + ……A + an + …… is a series.

If we take m terms of a finite sequence, the series can be written with summation notation
Σ (Sigma notation ).
m
X ak     = a1 + a2 + a3 + a4 + ……A + am
k=1
= The sum of all , the value of k being all the natural numbers from 1 to k,
k is called the summation variable A
4
1
Example 3 : Write in expanded form : X fffff  ffff
ffff
fffff
k=1 k + 1
3

1
fffff
ffff
ffff
fffff
Solution : Replace k in 3      with integers from 1 to 4 and add them
k +1
4
1           1       1       1         1          f f 1f 1f
1f 1f ff ff
fffff
ffff
ffff
X fffff = fffff fffff fffff fffff =
ffff fffff fffff fffff
ffff ffff
ffff ffff ffff
+ 3ffff+ ffff
+ 3                  ffff
+ + ff ff
f f ff ff
+
k=1 k + 1       1 +1 2 +1 3 +1 4 +1
3           3                         3
2 9 28 65
SERIES IDENTITIES:
n       n               n                                  n        n         n
X ak + X bk = X ak + bk                                 X ak @ X bk = X ak @ bk
b       c                                            b         c

k=1     k=1          k=1                                k=1         k=1     k=1
n               n                                       n
X cak = c X ak                                          X c = cn
k=1           k=1                                       k=1

nfff+fff2nffff
n f 1 ff + 1
n
`           a                                   n
`    a`            a
n@1
nfffffff
fffffff
ffffff
X k = fffffff                                                 fffffffffffff
fffffffffffff
X k = ffffffffffffff
2

k=1
2                                             k=1
6
n + 1 2n + 1 3n 2 + 3n @ 1
ab       c
n+1
n     2
`           a2                             n
`    a`
nfffffffffffffffffffffffff
nffffffff
ffffffff
fffffff
X k = ffffffff                                         Xk =           ffffffffffffffffffffffff
fffffffffffffffffffffffff
fffffffffffffffffffffffff
3                                                           4

k=1
4                                            k=1
30

ARITHMATIC SEQUENCE & SERIES
A sequence where we start with a number a and repeatedly add a fixed constant d to get
the subsequent numbers, is called an Arithmatic Sequence.

If we start with 2 and add a fixed constant 3 to it repeatedly, we get the sequence
2, 5, 8, 11, 14, 17, AAAAAAAAAAAA This is an arithmatic sequence.

An Arithmatic Sequence is always of the form
a, a + 2d, a + 3d, a + 4d, a + 5d, AAAAAAAAAAA
where we start with the number a and keep on adding d to get the subsequent terms.
The number a is the first term (can be written as a1 too) and d is the common
difference.

To get d in a given arithmetic sequence, subtract any term from its next term.
d = a2 @ a1 = a3 @ a2 = a4 @ a3 = ……AA = an @ an @ 1 = ……A

term of an arithmetic sequence is given by
The nth `
an = a + n @ 1 d     where an = nth term, a = a1 = first term, d = common difference A
a

If S n is the sum of n terms of a finite arithmetic sequence, then
nf       a nf
ff          ff
S n = f a + a n = f 2a + n @ 1 d
`           B    `      a C
2           2

Example : Write the first 5 terms of the arithmetic sequence 11,8,….. and find its 100th
term.

Solution : Here a1 = a = 11 and a2 = 8 A So d = a2 @ a1 = 8 @ 11 = @ 3 A
Thus each term can be found by adding -3 to the previous term.
Hence the first 6 terms are 11, 8, 5, 2, -1, -4.
For the 100th term n = 100, and we have already found a = 11, d = @ 3
an = a + n @ 1 d
`      a
Using
a100 = 11 + 100 @ 1 @ 3
`       a`   a

= 11 + 99 @ 3
`  a

= 11 @ 297
= @ 286

Example : Find the sum of first 12 terms of the arithmetic sequence given by
an = 2n + 5 A

Solution : Putting n=1 and 2 we get,
a = a1 = 2 B 1 + 5 = 7, a2 = 2 B 2 + 5 = 9 A   So d = a2 @ a1 = 9 @ 7 = 2 A
nf
f
ff
Using the summation formula S n = 2a + n @ 1 d
B      ` a C
2
12f
ff
f
f
ff
we get S 12 =     2 B7 + 12 @ 1 2 = 216 A
B      `      a C
2

GEOMETRIC SEQUENCE & SERIES :
A sequence where we multiply each term with a non-zero constant to get the next term, is
a Geometric Sequence.

If we start with 2 and multiply it repeatedly by a non-zero fixed number 3, we get the
2       3
sequence 2, 2 B3, 2 B3 , 2 B3 , AAAAAAAAAAAA or 2, 6, 18, 54, AAAAAAA This is a geometric
sequence.

A Geometric Sequence is always of the form
a, ar 2 , ar 3 , ar 4 , ar 5 AAAAAAAAAAAAAA
where we start with the number a and keep on multiplying with r to get the subsequent
terms.
The number a is the first term (can be written as a1 too) and r is the common ratio.

r is the ratio of any term to its previous term,
aff aff aff             anff
f 3
r = ff ff ff ……AA = ffff ……A
2f f 4
= f= f=  f            f
ffff
fff
=
a1 a2 a3               an @ 1

Examples of Geometric sequence :
1f 1f ff ff
f f 1f 1f
f f ff ff                                             1f     1f
, , ff ffAAAAAAAAA
, ,                                          f
a = f, r = f f
2 6 18 54                                              2      3
1, @ 10, 20, @ 40, 80, AAAAAAAAAA                  a = 1, r = @ 5
2, 0.2, 0.02, 0.002, AAAAAAAAAA                    a = 2, r = 0.1
The nth term of a Geometric Sequence is given by
an = a r n @ 1 where an = nth term, a = a1 = first term, r = common ratio A

A Geometric Series is the sum of the terms as indicated above. The sum of n terms of it
@rn
1fffff
fffff
fffff
is given by S n = a fffffwith r ≠ 1
1@r

Example : Write the 6th term and sum of first 6 terms of the geometric sequence
9ff
4, 6, f, AAAAAAAAA
2
Solution :
aff 6f 3f
f f f
f f
ff f f
Here in this geometric sequence, a = 4, and a2 = 6, the common ratio r = 2 = = . So,
a1 4 2
f g6 @ 1
3f
f
f         243f
using an = a r   n@1
,          6th term = a6 = 4 B                 = fff
ff

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