# Arithmetic Series

In an arithmetic series each term is increasing by a constant difference more than the previous term. The constant is called the common difference, d.

If ${ T }_{ 1 }, { T }_{ 2 } \& { T }_{ 3 }$ are consecutive terms of an arithmetic series then:

$d = { T }_{ 2 }-{ T }_{ 1 } = { T }_{ 3 }-{ T }_{ 2 }$

For example, we can use the arithmetic series $(k+2)+(3k+2)+(6k+2)+...$. What is the value of k so that the series exist?

${ T }_{ 2 }-{ T }_{ 1 }={ T }_{ 3 }-{ T }_{ 2 }\\ (3k+2)-(k+2)=(6k-1)-(3k+2)\\ 3k+2-k-2=6k-1-3k-2\\ 2k=3k-3\\ 0=k-3\\ k=3$

So, what is the common difference, d of the series?

To find the common difference we could find the first 3 terms of the series:

Substituting = 3:

${ T }_{ 1 }=k+2\\ { T }_{ 1 }=3+2=5\\ \\ { T }_{ 2 }=3k+2\\ { T }_{ 2 }=3(3)+2=11\\ \\ { T }_{ 3 }=6k-1\\ { T }_{ 3 }=6(3)-1=17$

So, the common difference, d = 11 - 5 = 17 - 11 = 6.

## Terms of an Arithmetic Series

$a+(a+d)+(a+2d)+(a+3d)+...+[a+(n-1)d]$ is an arithmetic series with first term a, common difference d and nth term given by:

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

### Examples

1. Find the 12th term of the series 2 + 6 + 10 + ...

= 2, d = 4, n = 12

${ T }_{ n }=a+(n-1)d\\ { T }_{ 12 }=2+(12-1)4\\ { T }_{ 12 }=46$

2. Find the first positive term of the series -50, -47, -44, ...

a = -50, d = 3

So for the first positive term, ${T}_{n} \ge 0$.

$a+(n-1)d \ge0 \\ -50+(n-1)3 \ge 0 \\ -50+3n-3 \ge 0 \\ 3n-53 \ge 0 \\ 3n \ge 53 \\ \\ n \ge 17\cfrac{2}{3} \\ \\ \therefore n=18 \ gives \ the \ first \ positive \ term$

${T}_{18}=-50+(18-1)3 \\ {T}_{18}=-30+17 \times 3 \\ {T}_{18}=1$

So the first positive term is 1.

3. The 5th term of an arithmetic series is 37 and the 8th term is 55. Find the common difference and the first term of the series.

## Partial Sum of an Arithmetic Series

The sum of the first n terms of an arithmetic series is given by the formula:

${ S }_{ n }=\cfrac { n }{ 2 } (a+l)$ where l = last or nth term

And if the nth term is unknown, we use the formula below in general:

${ S }_{ n }=\cfrac { n }{ 2 } [2a+(n-1)d]$

### Examples

1. Evaluate 10 + 15 + 20 + 25 + ... + 225.

a = 10, d = 5

First we find n.

${ T }_{ n }=a+(n-1)d\\ { T }_{ n }=225\\ 225=10+(n-1)5\\ 225=10+5n-5\\ 225=5n+5\\ 5n=220\\ n=44\\ \\ \\ { S }_{ n }=\cfrac { n }{ 2 } (a+l)\\ \\ { S }_{ 44 }=\cfrac { 44 }{ 2 } (10+225)\\ \\ { S }_{ 44 }=5170\\$

2. For what value of n is the sum of n terms of 2 + 11 + 20 + ... equal to 618?

$a=2,\quad d=9,\quad { S }_{ n }=618\\ \\ { S }_{ n }=\cfrac { n }{ 2 } [2a+(n-1)d]\\ \\ 618=\cfrac { n }{ 2 } [2\times 2+(n-1)9]\\ \\ 1236=n(4+9n-9)\\ 1236=n(9n-5)\\ 1236=9{ n }^{ 2 }-5n\\ 9{ n }^{ 2 }-5n-1236=0\\ (n-12)(9n+103)=0\\ \\ \therefore \quad n=12\quad or\quad n=-11.4$

But n cannot be negative, so n = 12.

3. Evaluate $\sum _{ r=1 }^{ 50 }{ 3r+2 }$

$\sum _{ r=1 }^{ 50 }{ 3r+2 } =(3\times 1+2)+(3\times 2+2)+...+(3\times 50+2)\\ \\ \qquad \qquad =5+8+...+152$
${ S }_{ n }=\cfrac { n }{ 2 } (a+l)\\ \\ { S }_{ 50 }=\cfrac { 50 }{ 2 } (5+152)\\ \\ { S }_{ 50 }=25\times 157\\ \\ { S }_{ 50 }=3925$