Which term of the G.P. 2 , ? 1 , ? 1 2 , ? 1 4 , ... is 1 128 ?

Question

Which term of the G.P. $2, 1, \frac{1}{2}, \frac{1}{4}$ , ... is $\frac{1}{128}$ ?

Smart Achievers · Accepted Answer

(1)

$Here, first term (a) = 2 and common ratio (r) = \frac{1}{2}$

$Let the n^{th} term be \frac{1}{128} . Then,$

$a_{n} = \frac{1}{128}$ $\Rightarrow a r^{n - 1} = \frac{1}{128}$ $\Rightarrow 2 {(\frac{1}{2})}^{n - 1} = \frac{1}{128}$

$\Rightarrow$ ${(\frac{1}{2})}^{n - 1} = \frac{1}{2^{8}}$ $\Rightarrow {(\frac{1}{2})}^{n - 1} = {(\frac{1}{2})}^{8}$

$\Rightarrow$ $n - 1 = 8$ $\Rightarrow n = 9$

$Thus, 9^{th} term of the given G.P. is \frac{1}{128}$

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