An algebra problem by Aparna Kalbande

Algebra Level 2

The first four terms of a geometric progression are 5 , 10 , 20 , 40 5,10,20,40 and consider 1280 1280 is its n th n^{\text{th}} term, then what is n n ?

None of the others 10 9 8

Aparna Kalbande by Aparna Kalbande , 20, India

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1 solution

Common ratio= 10 5 = 2 \dfrac{10}{5}=2
Since nth term of G.P. is a r n − 1 ar^{n-1} . [Here 'a' is first term and 'r' is common ratio]
Given, 1280 = a r n − 1 1280=ar^{n-1}
1280 = 5 × 2 n − 1 1280=5×2^{n-1}
256 = 2 n − 1 256=2^{n-1}
2 8 = 2 n − 1 2^8=2^{n-1}
8 = n − 1 8=n-1
n = 9 n=\boxed{9} .

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