Which term of the sequence 2, 2 2, 4, .... is 128? - Vidyadip

Question

Which term of the sequence $2, 2\sqrt 2, 4, .... is 128?$

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Answer

Here $a = 2, r = \frac { 2 \sqrt { 2 } } { 2 } = \sqrt { 2 }$ and $a_n = 128$
$\therefore a_n = ar^{n-1}$
$\Rightarrow 128 = 2 \times ( \sqrt { 2 } ) ^ { n - 1 }$
$\Rightarrow 64 = ( \sqrt { 2 } ) ^ { n - 1 }$
$\Rightarrow ( \sqrt { 2 } ) ^ { 12 } = ( \sqrt { 2 } ) ^ { n - 1 }$
$\Rightarrow n - 1 = 12$
$\Rightarrow n = 13$
Therefore, $13^{th} $ term of the given G.P. is $128$

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