Find the sum to n terms in the geometric progression 1, -a, a², -a³ …. (if a ≠ -1)

Solution:

The given G.P is

1, - a, a², - a³, ....

Here, a₁ = 1 and r = - a

It is known that S_n = a (1 - rⁿ)/(1 - r)

Therefore,

S_n = 1 (1 - (- a)ⁿ)/(1 - (- a))

= [1 - (- a)ⁿ]/(1 + a)

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 9

 Find the sum to n terms in the geometric progression 1, -a, a², -a³ …. (if a ≠ -1)

Summary:

We know that the sum of a G.P is S_n = a (1 - rⁿ)/(1 - r) and therefore the sum of n terms of the G.P is [1 - (- a)ⁿ]/(1 + a)