Question

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

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

Answer 1

The correct option is C

$\frac{2}{5}$

Explanation For The Correct Option:

Finding the $5th$ term of the series:

The given series,

$\frac{10}{9},\left(\frac{1}{3}\right)\left(\sqrt{\frac{20}{3}}\right),\frac{2}{3},......$

Now dividing the second term by the first term

$$\begin{gathered} \Rightarrow \left(\frac{1}{3}\right)\sqrt{\frac{20}{3}}÷\frac{10}{9}=\left(\frac{1}{3}\right)\sqrt{\frac{20}{3}}\times \frac{9}{10} \\ =\sqrt{\frac{3}{5}} \end{gathered}$$

Now dividing the third term by the second term

$$\begin{gathered} \Rightarrow \frac{2}{3}÷\left(\frac{1}{3}\right)\left(\sqrt{\frac{20}{3}}\right)=\left(\frac{1}{3}\right)\left(\sqrt{\frac{20}{3}}\right)\times \frac{3}{2} \\ =\sqrt{\frac{3}{5}} \end{gathered}$$

Hence this is a GP with common ratio $r=\sqrt{\frac{3}{5}}$ and first term $a=\frac{10}{9}$

Since the $nth$ term of GP $=a{r}^{n-1}$

Therefore, $5th$ term of this GP

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

Hence, option (C) is the correct answer.