Question

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

• A. 10th term
• B. 8th term
• C. 9th term
• D. 11th term

Answer 1

Answer: (B)

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
$n-1=7$
$\Rightarrow$ $n=8$