Question

the terms of a G.P. with first term a and common ratio r are squared. prove that resulting numbers form a G.P..find its first term, common ratio and the nth term.

¤ (chapter -arithmetic ans geometric progression)

¤please answer fast. .....

¤15 points! !!!

Answer 1

Hey there !!!!

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Let sequence of Geometric Progression be :

=a,ar,ar²,ar³.................arⁿ⁻¹

According to question each term of given series is squared 

a becomes a² 
ar becomes a²r²
......................... arⁿ⁻¹ becomes (arⁿ⁻¹)² =a²r²ⁿ⁻²

So new series is

a²,a²r²,a²r⁴.................a²r²ⁿ⁻²

First term = a²

Common ratio = a²r²/a² = r²

nth term = a²r²ⁿ⁻²

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EXAMPLE :

Let series be 2,4,8,16............

With 1st term =2 and common ratio = 2

Squaring each term of series

=2²,4²,8²,16²

4,16,64,256

Common ratio = 16/4 = 4

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Hope this helped you.............

Answer 2

Step-by-step explanation:

Let AP be a,ar,ar^2,ar^3........ar^n-1

When these terms are squared then resulting GP would be a^2,a^2r^2,a^2r^4,a^2r^6,........,a^2r^2(n-1)

Therefore it's first term would be a^2,common ratio will be r^2 and nth term would be a^2r^2(n-1)

Happy to help!