Sequence and Series

Sequence and Series:

In this article, read all you need to know about sequence and series.

Introduction

An arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. 

2,4,6,8,10….is an arithmetic sequence with the common difference 2.

A sequence may be named or referred to by an upper-case letter such as “A” or “S“. The terms of a sequence are usually named something like “a_i” or “a_n“, with the sub-scripted letter “i” or “n” being the “index” or the counter. So the second term of a sequence might be named “a₂“, and “a₁₂” would designate the twelfth term.

If the first term of an arithmetic sequence is a₁ and the common difference is d, then the nth term of the sequence is given by:

1. A series in which the difference between any term and its previous term is constant (eg. 2 + 5 + 8 + 11 + 14) is an arithmetic series.
2. A sequence T_n is said to be an arithmetic progression (A . P) if there exists a number, say d, such that Tn + 1 − Tn = d, n = 1. The constant number d is called the common difference of the corresponding A . P
3. If a and d be the first term and common difference of the A . P, then T_n is said to be the last term of the series and the formulafor calculation is Tn = a + (n − 1)d, where n ∈ N.
4. If a and d be the first term and common difference of the A . P. Tn, then the sum of first n terms, Sn is given by Sn = n2 [2a + (n − 1)d ], n ∈ N.
5. T1 = S1 and for n > 1, we have Tn = Sn − Sn − 1
6. If the sequence a, A1, A2, ....., An, b is an A . P., then the numbers A1, A2, ....., An are called the n arithmetic means between a and b.
7. The A.M. between given numbers a and b is equal to (a + b)/2.
8. The sum of n A.M.s between given numbers a and b is equal to n times the A.M between a and b.
9. If a, b, c are in A.P., then for any k, a + k, b + k, c + k are in A.P. a − k, b − k, c − k are in A.P. ka, kb, kc are in A.P. a/k, b/k, c/k are in A.P. (k¹ 0)
10. If the sum of three numbers in A.P. is given, then the numbers should be taken as a − d, a, a + d. If the sum of four numbers in A.P. is given, then the numbers should be taken as a − 3d, a − d, a + d, and a + 3d.

Examples

Example 1: Find 10^th term in the series 1, 3, 5, 7, …

Solution: 

a = 1
d = 3 – 1 = 2

10^th term, t₁₀ = a + (n-1)d = 1 + (10 – 1)2 = 1 + 18 = 19

Example 2: Find 16^th term in the series 7, 13, 19, 25, …

Solution: 

a = 7
d = 13 – 7 = 6

16^th term, t₁₆ = a + (n-1)d = 7 + (16 – 1)6 = 7 + 90 = 97

Example 3: Find the number of terms in the series 8, 12, 16, . . .72

Solution:

a = 8
l = 72
d = 12 – 8 = 4

$n=[\frac{(l-a)/}{\mathrm{d]}}+1=\frac{(72-8)}{4}+1=\frac{64}{4}+1=16+1=17$

Example 4: Find 4 + 7 + 10 + 13 + 16 + . . . up to 20 terms

Solution: 

a = 4
d = 7 – 4 = 3

Sum of first 20 terms, S₂₀
$=\frac{\mathrm{n/}}{2}[2a+(n-1\left)d\right]=\frac{\mathrm{20/}}{2}\left[\right(2\times 4)+(20-1\left)3\right]=10[8+(19\times 3\left)\right]=10(8+57)\; =\; 650$

Example 5: Find 6 + 9 + 12 + . . . + 30

Solution: 

a = 6
l = 30
d = 9 – 6 = 3

$n=[\frac{(l-a)/}{\mathrm{d]}}+1=\; [\frac{(30-6)/}{\mathrm{3]}}+1=\frac{24}{3}+1=8+1=9$

Sum S $=\frac{\mathrm{n/}}{2}(a+l)=\frac{\mathrm{9/}}{2}(6+30)=\frac{9}{2}\times 36=9\times 18=162$

Arithmetic Mean

If a, b, c are in AP, b is the Arithmetic Mean (AM) between a and c. In this case, $b=\frac{\mathrm{1/}}{2}(a+c)$

If a, a₁, a₂ … a_n, b are in AP we can say that a₁, a₂ … a_n are the n Arithmetic Means between a and b.

To solve most of the problems related to AP, the terms can be conveniently taken as
3 terms: (a – d), a, (a +d)
4 terms: (a – 3d), (a – d), (a + d), (a +3d)
5 terms: (a – 2d), (a – d), a, (a + d), (a +2d)

In an AP, sum of terms equidistant from beginning and end will be constant.

Harmonic Progression

Non-zero numbers $a_1,a_2,a_3,\cdots a_n$ are in Harmonic Progression(HP) if $\frac{\mathrm{1/}}{a_1},\frac{\mathrm{1/}}{a_2},\frac{\mathrm{1/}}{a_3},\cdots \frac{\mathrm{1/}}{a_n}$ are in AP. Harmonic Progression is also known as harmonic sequence.

Three non-zero numbers a, b, c will be in HP, if  $\frac{\mathrm{1/}}{a},\frac{\mathrm{1/}}{b},\frac{\mathrm{1/}}{c}$ are in AP

The Harmonic Mean(HM) between two numbers a and b = $\frac{2a\mathrm{b/(}}{a+\mathrm{b)}}$

If a, b, c are in HP, $\frac{2}{b}=\frac{\mathrm{1/}}{a}+\frac{\mathrm{1/}}{c}$

Geometric Progression

Geometric Progression(GP) or Geometric Sequence is sequence of non-zero numbers in which the ratio of any term and its preceding term is always constant.

A geometric progression(GP) is given by a, ar, ar², ar³, …  [where a = the first term , r = the common ratio]

Sum of an Infinite Geometric Progression

$S_{\infty }=\frac{\mathrm{a/(}}{1-\mathrm{r)}}\text{if -1 < r < 1}$

where a= the first term , r = common ratio

Example: Find $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \infty$

Solution: 

a = 1,       r = $\frac{(\frac{\mathrm{1/}}{2})/}{1}=\frac{\mathrm{1/}}{2}$

Here -1 < r < 1. Hence,
$S_{\infty }=\frac{\mathrm{a/(}}{1-\mathrm{r)}}=\frac{\mathrm{1/}}{(1-\frac{\mathrm{1/}}{2})}$

$=\frac{\mathrm{1/}}{\left(\frac{\mathrm{1/}}{2}\right)}=2$

Power Series – Important Formulas

$1+1+1+\cdots \text{n terms}$ $=\sum 1=n$

$1+2+3+\cdots +n$$=\sum n=\frac{n(n+1)/}{2}$

$1^2+2^2+3^2+\cdots +n^2$ $=\sum n^2=\; [\frac{n(n+1)(2n+1)]/}{6}$

$1^3+2^3+3^3+\cdots +n^3$ $=\sum n^3=\frac{n^2(n+1{)}^{\mathrm{2/}}}{4}={\left[\frac{n(n+1)/}{2}\right]}^2$