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Q.
Which term of the series $ \sqrt{2},\frac{2}{3},\,2\frac{\sqrt{2}}{9},.... $ is $ \frac{16}{2187} $ ?
J & K CETJ & K CET 2014Sequences and Series
Solution:
Given, series is $ \sqrt{2},\frac{2}{3},\frac{2\sqrt{2}}{9},.... $ or $ \frac{\sqrt{2}}{{{3}^{0}}},\frac{{{(\sqrt{2})}^{2}}}{3},\,\frac{{{(\sqrt{2})}^{3}}}{{{3}^{2}}},.... $
which is an infinite GP whose, first term
$ a=\frac{\sqrt{2}}{{{3}^{0}}}=\sqrt{2} $
and common ratio
$=\frac{{{(\sqrt{2})}^{2}}}{3}\times \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{3} $
Let nth term of GP
$=\frac{16}{2187} $
$ \therefore $ $ a{{t}^{n-1}}=\frac{16}{2187} $
$ \Rightarrow $ $ \sqrt{2}{{\left( \frac{\sqrt{2}}{3} \right)}^{n-1}}=\frac{16}{2187} $
$ \Rightarrow $ $ {{\left( \frac{\sqrt{2}}{3} \right)}^{n-1}}=\frac{8\sqrt{2}}{2187}={{\left( \frac{\sqrt{2}}{3} \right)}^{7}} $
On comparing the powers, we get
$ n-1=7 $
$ \Rightarrow $ $ n=8 $