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 + ...

Show Answer

= 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, ...

Show Answer

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.

Show Answer

 

 

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.

Show Answer

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?

Show Answer

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 }

 

Show Answer

\sum _{ r=1 }^{ 50 }{ 3r+2 } =(3\times 1+2)+(3\times 2+2)+...+(3\times 50+2)\\ \\ \qquad \qquad =5+8+...+152

Arithmetic series with a = 5, d = 3, l = 152, n = 50

{ S }_{ n }=\cfrac { n }{ 2 } (a+l)\\ \\ { S }_{ 50 }=\cfrac { 50 }{ 2 } (5+152)\\ \\ { S }_{ 50 }=25\times 157\\ \\ { S }_{ 50 }=3925