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# Find the sum to n terms in the geometric progression 1, -a, a^{2}, -a^{3} …. (if a ≠ -1)

**Solution:**

The given G.P is

1, - a, a^{2}, - a^{3}, ....

Here, a_{1} = 1 and r = - a

It is known that S_{n} = a (1 - r^{n})/(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, a^{2}, -a^{3} …. (if a ≠ -1)

**Summary:**

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

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