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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