Question

The $5th$ term of the series$\frac{10}{9},\frac{1}{3}\sqrt{\frac{20}{3}},\frac{2}{3}$ is

• A. $\frac{1}{3}$
• B. 1
• C. $\frac{2}{5}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

The correct option is C

$\frac{2}{5}$

Explanation for the correct answer :

Step-1 : To check whether the given series in $G.P.$

Formula to be used : We know that three numbers $a,b,c$ are in $G.P.$ if $b^2=ac$.

The first three terms of the given series are : $a=\frac{10}{9}$, $b=\frac{1}{3}\sqrt{\frac{20}{3}}$ and $c=\frac{2}{3}$.

So, we get

$$\begin{gathered} b^2={\left(\frac{1}{3}\sqrt{\frac{20}{3}}\right)}^2 \\ =\frac{20}{27} \\ =\frac{10}{9}\times \frac{2}{3} \\ =ac \end{gathered}$$

Thus, the given series is a geometric progression.

Step-2 : Finding the common ratio

Let the common ration of the given $G.P.$ be $r$. Then we should have

$$\begin{gathered} r=\frac{b}{a} \\ =\frac{{\displaystyle \frac{1}{3}}\sqrt{{\displaystyle \frac{20}{3}}}}{{\displaystyle \frac{10}{9}}} \\ =\frac{\sqrt{20}}{3\sqrt{3}}\times \frac{9}{10} \\ =\sqrt{\frac{6}{10}} \\ =\sqrt{\frac{3}{5}} \end{gathered}$$

Step-3 : Finding the $5th$term of the series

Formula to be used : We know that the $nth$ term of a $G.P.$ with the first term $a$ and the common ratio $r$ is $a{r}^{n-1}$.

Here, $a=\frac{10}{9}$ and $r=\sqrt{\frac{3}{5}}$. So, the $5th$ term will be

$$\begin{gathered} t_5=a{r}^{5-1} \\ =\frac{10}{9}\times {\left(\sqrt{\frac{3}{5}}\right)}^4 \\ =\frac{10}{9}\times \frac{9}{25} \\ =\frac{2}{5} \end{gathered}$$

Hence, the correct answer is option (C)