Question

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

Answer 1

Given:

A G.P.↠ 1,-a,a²,-a³......to n terms

To Find:

Sum to n terms of the given G.P.

Solution:

We know that,

Sum of first n terms of G.P. is given by

$\pink{\boxed{\mathrm{S_{n}=\dfrac{a(1-r^{n})}{1-r},\mid r\mid<1}}}$

$\green{\boxed{\mathrm{S_{n}=\dfrac{a(r^{n}-1)}{r-1},\mid r\mid>1}}}$

where,

a is first term of G.P.

r is common difference of G.P.

n is no. of terms of G.P.

Now, for the given G.P.

First term,a= 1

Common difference,r= -a

So,

Sum for given G.P. is

↬ When |a|<1

Case-1: If n is even

$\mathrm{S_{n}=\dfrac{1(1-(-a)^{n})}{1-(-a)}}$

$\mathrm{S_{n}=\dfrac{1-a^{n}}{1+a}}$

Case-2: If n is odd

$\mathrm{S_{n}=\dfrac{1(1-(-a)^{n})}{1-(-a)}}$

$\mathrm{S_{n}=\dfrac{1-(-a^{n})}{1+a}}$

$\mathrm{S_{n}=\dfrac{1+a^{n}}{1+a}}$

↬ When |a|>1

Case-1: If n is even

$\mathrm{S_{n}=\dfrac{1((-a)^{n}-1)}{-a-1}}$

$\mathrm{S_{n}=\dfrac{a^{n}-1}{-a-1}}$

$\mathrm{S_{n}=\dfrac{a^{n}-1}{-(a+1)}}$

$\mathrm{S_{n}=\dfrac{-(a^{n}-1)}{a+1}}$

$\mathrm{S_{n}=\dfrac{1-a^{n}}{a+1}}$

Case-2: If n is odd

$\mathrm{S_{n}=\dfrac{1((-a)^{n}-1)}{-a-1}}$

$\mathrm{S_{n}=\dfrac{-a^{n}-1}{-a-1}}$

$\mathrm{S_{n}=\dfrac{-(a^{n}+1)}{-(a+1)}}$

$\mathrm{S_{n}=\dfrac{a^{n}+1}{a+1}}$