Question

# Which term of the series $\sqrt{2},\frac{2}{3},\frac{2\sqrt{2}}{9},......$... is $\frac{16}{2187}$?

A

${10}^{th}$term

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B

${8}^{th}$term

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C

${9}^{th}$term

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D

${11}^{th}$term

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Solution

## The correct option is B ${8}^{th}$termStep 1: Note the given dataThe common ratio of a GP series is $\frac{{a}_{n}}{{a}_{n-1}}$.Where ${a}_{n}$is a ${n}^{th}$term of the GP and ${a}_{n-1}$is a ${\left(n-1\right)}^{th}$term of the GP.The series is $\sqrt{2},\frac{2}{3},\frac{2\sqrt{2}}{9},......$The ratio of the second term to the first term is $\frac{\frac{2}{3}}{\sqrt{2}}=\frac{\sqrt{2}}{3}$.The ratio of the third term to the second term is $\frac{\frac{2\sqrt{2}}{9}}{\frac{2}{3}}=\frac{\sqrt{2}}{3}$.The given series is in geometric progression the common ratio $r=\frac{\sqrt{2}}{3}$.$a=\sqrt{2};{a}_{n}=\frac{16}{2187}$Step 2: Find the value of $n$Geometric progression is a series of numbers where the ratio of two consecutive terms remains constant.The formula to find the ${n}^{th}$term of a geometric progression is ${a}_{n}=a{r}^{n-1}$$⇒$$\frac{16}{2187}=\sqrt{2}×{\left(\frac{\sqrt{2}}{3}\right)}^{n-1}$$⇒$$\frac{16}{2187}={\frac{\left(\sqrt{2}\right)}{{\left(3\right)}^{n-1}}}^{n}$$⇒$$\frac{16}{2187}={\frac{3\left(\sqrt{2}\right)}{{\left(3\right)}^{n}}}^{n}$$⇒$$\frac{16}{6561}={\frac{\left(\sqrt{2}\right)}{{\left(3\right)}^{n}}}^{n}$$⇒$$\frac{{2}^{4}}{{3}^{8}}=\frac{{\left(2\right)}^{\frac{1}{2}n}}{{\left(3\right)}^{n}}$$⇒$$n=8\left[\because \mathrm{if}{a}^{m}={a}^{n},\mathrm{then}m=n\right]$Hence, the correct option is B.

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