Question

In a G.P., $t_2=\frac{3}{5}$ and $t_3=\frac{1}{5}$. Then the common ratio is

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

Answer 1

Answer: (B)

$\frac{1}{3}$

We know that the general term of an geometric progression with first term $a$ and common ratio $r$ is $T_n=a{r}^{n-1}$

It is given that the $T_2=\frac{3}{5}$ that is:

$T_2=a{r}^{2-1}\Rightarrow \frac{3}{5}=ar......\left(1\right)$

Also it is given that $T_3=\frac{1}{5}$ that is:

$T_3=a{r}^{3-1}\Rightarrow \frac{1}{5}=a{r}^2\Rightarrow \frac{1}{5}=ar\times r\Rightarrow \frac{1}{5}=\frac{3}{5}\times r\left(using\right(1\left)\right)\Rightarrow 3r=1\Rightarrow r=\frac{1}{3}$

Hence, the common ratio is $\frac{1}{3}$.