Question

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

• A. ${10}^{th}$term
• B. $8^{th}$term
• C. $9^{th}$term
• D. ${11}^{th}$term

Answer 1

The correct option is B

$8^{th}$term

Step 1: Note the given data

The 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 $\frac{{\displaystyle \frac{2}{3}}}{\sqrt{2}}=\frac{\sqrt{2}}{3}$.

The ratio of the third term to the second term is $\frac{{\displaystyle \frac{2\sqrt{2}}{9}}}{{\displaystyle \frac{2}{3}}}=\frac{\sqrt{2}}{3}$.

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 $a_n=a{r}^{n-1}$

$\Rightarrow$$\frac{16}{2187}=\sqrt{2}\times {\left(\frac{\sqrt{2}}{3}\right)}^{n-1}$

$\Rightarrow$$\frac{16}{2187}={\frac{\left(\sqrt{2}\right)}{{\left(3\right)}^{n-1}}}^n$

$\Rightarrow$$\frac{16}{2187}={\frac{3\left(\sqrt{2}\right)}{{\left(3\right)}^n}}^n$

$\Rightarrow$$\frac{16}{6561}={\frac{\left(\sqrt{2}\right)}{{\left(3\right)}^n}}^n$

$\Rightarrow$$\frac{2^4}{3^8}=\frac{{\left(2\right)}^{{\displaystyle \frac{1}{2}}n}}{{\left(3\right)}^n}$

$\Rightarrow$$n=8\left[\because \mathrm{if}{a}^m=a^n,\mathrm{then}m=n\right]$

Hence, the correct option is B.