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Q. Which term of the GP $ 3,3\sqrt{3},9.... $ is $ 2187 $ ?

KEAMKEAM 2004

Solution:

Let nth term of GP is 2187 $ \therefore $
$ 3{{(\sqrt{3})}^{n-1}}=2187 $
$ \Rightarrow $ $ {{3}^{(n/2-1/2+1)}}={{3}^{7}} $
$ \Rightarrow $ $ \frac{n}{2}+\frac{1}{2}=7 $
$ \Rightarrow $ $ n=13 $