# Arithmetico Geometric Series

A series is said to be in arithmetico geometric series if each of its term is the product of the corresponding terms of an AP and a GP.

For example, 1, 2

*x*, 3*x*^{2}, 4*x*^{3},…In the above series, the first part of this series is in an AP (1, 2, 3, 4,…) and the 2nd part is (

*x*^{0},*x*^{1},*x*^{2},*x*^{3},….) in a GP.Sum of n terms of any arithmetico geometric series (AGS)

The sum of

*n*terms of any AGS a, (*a*+*d*)r, (*a*+ 2*d*)r^{2},…is given by.

*r*^{n }^{, }if*r*≠ 1.S

_{n}=[2*a*+ (*n*− 1)*d*], if*r*= 1Sum of infinite terms of any arithmetico geometric series (AGS)

However, I would suggest students to desist from using these formulae. They should use the standard process to find out the sum of any AGS which is given below:

*N*be the sum of the arithmetico geometric series. Then each term of the series is multiplied by

*r (the common ratio of GP) and is written by shifting each term one step rightward, and then by subtracting*

*r*

*N*from

*N*to get (1−r)

*N*. Thus

*N*is finally obtained.

Example-1

What is the sum of the following series till infinity: 1 + 2

*x*+ 3*x*^{2}+ 4*x*^{3}+ …, |*x*| <1Solution

Assume S = 1 + 2

*x*+ 3*x*^{2}+ 4*x*^{3}+…(1)Multiplying S by

*x*,*x*. S =*x*+ 2*x*^{2}+ 3*x*^{3}+ 4*x*^{4}+ …(2)Subtracting (2) from (1)

S –

*x*S = 1 + (*x*+*x*^{2}+*x*^{3}+…∝)S(1 −

*x*) = 1 + (*x*+*x*^{2}+*x*^{3}+…∝) =