# Series

Contents

### Summary

• When the terms of a sequence are added together a series is formed.
• A series can be denoted by a sigma notation  $\sum _{ a }^{ b }{ x }$  where a is the first value of n terms, b is the last value of the n terms and x is the expression of the given sequence.
• The formula for n-th term  ${ T }_{ n }$  of an arithmetic progression is:  ${ T }_{ n }\quad =\quad a\quad +\quad (n\quad -\quad 1)d$
• Sum  ${ S }_{ n }$  of the series is given by the formula:  ${ S }_{ n }\quad =\quad \frac { n }{ 2 } \left[ 2a\quad +\quad (n\quad -\quad 1)d \right]$
• The formula for n-th term  ${ T }_{ n }$  of a geometric progression is:  ${ T }_{ n }\quad =\quad a{ r }^{ n\quad -\quad 1 }$
• Formula to work out the sum of geometric progression:  ${ S }_{ n }\quad =\quad \frac { a(1\quad -\quad { r }^{ n }) }{ 1\quad -\quad r }$
• Sum to infinity of the geometric progression:  ${ S }_{ \infty }\quad =\quad \frac { a }{ 1\quad -\quad r }$

Let’s just quickly recall that sequence is a set of numbers in a given order with a rule for obtaining the terms. Moving on, when the terms of a sequence are added together a series is formed.

For example:

SequenceSeries
{1, 3, 5, 7….}1 + 3 + 5 + 7 + ....
{5, 10, 15, 20….}5 + 10 + 15 + 20 + ....
{10, 100, 1000,.....}10 + 100 + 1000 +.....

A series can be denoted by a sigma notation  $\sum _{ a }^{ b }{ x }$   where a is the first value of n terms, b is the last value of the n terms and x is the expression of the given sequence. This sigma notation is a Greek capital and is used to represent a sum.

Let’s consider we are given a series 2 + 4 + 6 + 8 + 10.

We can see that n = 5 is the last value of n and the expression of the sequence would be  $2\quad \times \quad n$.

Therefore, we can write it using the sigma notation in the form:

${ S }_{ n }\quad =\quad \sum _{ n\quad =\quad 1 }^{ 5 }{ 2n }$

Next, to find the sum of the above sequence, we carry out the following steps:

${ S }_{ n }\quad =\quad \sum _{ n\quad =\quad 1 }^{ 5 }{ 2n } \quad =\quad 2(1)\quad +\quad 2(2)\quad +\quad 2(3)\quad +\quad 2(4)\quad +\quad 2(5)$
$=\quad 2\quad +\quad 4\quad +\quad 6\quad +\quad 8\quad +\quad 10\quad$
$=\quad 30$

Furthermore, in this article we will study series in the following types of sequences.

#### Arithmetic Progression

An arithmetic progression is a sequence of numbers which increases by a constant amount which could be either positive or negative. This amount is called the ”common difference” (d) and the starting number is called ”first term” (a).

In general, if an arithmetic progression has first term a and common difference d, then in standard form, it is written as:

$a,\quad (a\quad +\quad d),\quad (a\quad +\quad 2d),\quad (a\quad +\quad 3d)\quad +......$

So the formula for n-th term ${ T }_{ n }$ of an arithmetic progression is:

${ T }_{ n }\quad =\quad a\quad +\quad (n\quad -\quad 1)d$

Here: a = first term, d = common differencen = no. of terms${ T }_{ n }$  = n-th term

#### Example #1

Q. Find the 20 th term of the sequence 5, 8, 11, 14, 17,…..

Solution:

We know a = 5, d = 8 – 5 = 3 also, we know it is an arithmetic sequence

${ T }_{ n }\quad =\quad a\quad +\quad (n\quad -\quad 1)d$
${ T }_{ 20 }\quad =\quad 5\quad +\quad (20\quad -\quad 1)(3)$
${ T }_{ 20 }\quad =\quad 5\quad +\quad 19(3)$
${ T }_{ 20 }\quad =\quad 5\quad +\quad 57$
${ T }_{ 20 }\quad =\quad 62$       Ans

#### Sum of Arithmetic Progression

Suppose we are asked to find the sum of first 100 numbers in an arithmetic progression. Adding all the numbers one by one will be a very tedious job. Gauss, a german mathematics solved this problem in a very simple manner.

If the first term is a, and the last term is l, the number of terms is n, then the sum ${ S }_{ n }$ of the series is given by the formula:

${ S }_{ n }\quad =\quad \frac { (a\quad +\quad 1)n }{ 2 }$

Or

${ S }_{ n }\quad =\quad \frac { n }{ 2 } \left[ 2a\quad +\quad (n\quad -\quad 1)d \right]$

#### Example #2

Q. Find the sum of the first 10 terms of the sequence 5, 8, 11, 14, 17,…..

Solution:

We know a = 5, d = 3 and n = 10. We know it is an arithmetic sequence.

${ S }_{ n }\quad =\quad \frac { n }{ 2 } \left[ 2a\quad +\quad (n\quad -\quad 1)d \right]$

${ S }_{ 10 }\quad =\quad \frac { 10 }{ 2 } \left[ 2(5)\quad +\quad (10\quad -\quad 1)(3) \right]$

${ S }_{ 10 }\quad =\quad 5\left[ 10\quad +\quad 27 \right]$

${ S }_{ 10 }\quad =\quad 5\left[ 37 \right] \quad =\quad 185$       Ans

#### Geometric Progression

Geometric progression is a sequence in which each term is multiplied by a constant r known as the common ratio. Thus if a is the first term, r is the common ratio, then the standard form of a geometric progression is:

$a,\quad a\quad \times \quad r,\quad a\quad \times \quad { r }^{ 2 },\quad a\quad \times \quad { r }^{ 3 },\quad a\quad \times \quad { r }^{ 4 },\quad ......$

Hence to find the term of the geometric progression, we use the formula:

${ T }_{ n }\quad =\quad a{ r }^{ n\quad -\quad 1 }$

Where r can be worked out by dividing two consecutive terms in such a way:

$r\quad =\quad \frac { { T }_{ 2 } }{ { T }_{ 1 } } \quad or\quad r\quad =\quad \frac { { T }_{ 3 } }{ { T }_{ 2 } } \quad or\quad r\quad =\quad \frac { { T }_{ 4 } }{ { T }_{ 3 } }$

#### Example #3

Q. Find the 10th term of the geometric progression 3, 6, 12,….

Solution:

We know a = 3, n = 10 and we can find r

$r\quad =\quad \frac { 6 }{ 3 }$
$r\quad =\quad 2$

Use the above formula to work out the 10th term.

${ T }_{ 10 }\quad =\quad 3{ (2) }^{ 10\quad -\quad 1 }$
${ T }_{ 10 }\quad =\quad 3{ (2) }^{ 9 }$
${ T }_{ 10 }\quad =\quad 3{ (512) }$
${ T }_{ 10 }\quad =\quad 1536$      Ans

#### Sum of geometric progression

Just like we studied a formula to calculate the sum of arithmetic progression. Similarly, we are given a formula to work out the sum of geometric progression:

${ S }_{ n }\quad =\quad \frac { a(1\quad -\quad { r }^{ n }) }{ 1\quad -\quad r }$

Here a = first term, r = common ratio, n = no. of terms

#### Sum to infinity of a geometric progression

Consider the following infinite geometric progression:

$1\quad +\quad \frac { 1 }{ 3 } \quad +\quad \frac { 1 }{ 9 } \quad +\quad \frac { 1 }{ 27 } \quad +\quad .......$

We can see that a = 1 and $r\quad =\quad \frac { 1 }{ 3 }$,

${ S }_{ n }\quad =\quad \frac { 1(1\quad -\quad { ({ 1 }/{ 3 }) }^{ n }) }{ (1\quad -\quad ({ 1 }/{ 3 })) }$
$=\quad \frac { 3 }{ 2 } \left[ { 1\quad -\quad (\frac { 1 }{ 3 } ) }^{ n } \right]$
$=\quad \frac { 3 }{ 2 } (1\quad -\quad \frac { 1 }{ { 3 }^{ n } } )$

As n increases  $\frac { 1 }{ { 3 }^{ n } }$  decreases, when n becomes very large i.e  $n\quad \rightarrow \quad \infty$,   $\frac { 1 }{ { 3 }^{ n } }$  becomes very small and  $\frac { 1 }{ { 3 }^{ n } } \quad \rightarrow \quad 0$.  The value of ${ S }_{ n }$ becomes  $\frac { 3 }{ { 2 }$.

We call the limiting value of  ${ S }_{ n }$  as the sum to infinity of the geometric progression.

Therefore, if we consider -1 <  r  <  1, then  ${ r }^{ n }\quad \rightarrow \quad 0$  when  ${ n }\quad \rightarrow \quad \infty$.

Hence the formula becomes:

${ S }_{ \infty }\quad =\quad \frac { a }{ 1\quad -\quad r }$

#### Example #4

Q. Find the sum to infinity of the geometric progression:

$2\quad +\quad \frac { 1 }{ 2 } \quad +\quad \frac { 1 }{ 8 } \quad +\quad \frac { 1 }{ 32 } \quad +\quad .......$

Solution:

a = 2 and $r\quad =\quad \frac { 1 }{ 4 }$

Using the above formula:

${ S }_{ \infty }\quad =\quad \frac { a }{ 1\quad -\quad r }$

$=\quad \frac { 2 }{ 1\quad -\quad \frac { 1 }{ 4 } }$

$=\quad 2\quad \div \quad \frac { 3 }{ 4 }$

$=\quad \frac { 8 }{ 3 }$    Ans