Question

In a geometric progression, the ratio of the sum of the first three terms and the first six terms is $125:152$. The common ratio.

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{4}{5}$
• D. $\frac{3}{5}$

Answer 1

The correct option is D

$\frac{3}{5}$

Explanation for the correct option:

Find the common ratio of GP:

Given, that the ratio of the sum of the first three terms to the sum of first six terms of a GP is $125:152$

We know the formula for the sum of $n$ terms of a GP is $S_n=\frac{a\left(r^n-1\right)}{r-1}$

So,

Sum of first $3$ terms of GP $S_3=\frac{a\left(r^3-1\right)}{r-1}$

Sum of first $6$ terms of GP $S_6=\frac{a\left(r^6-1\right)}{r-1}$

Now,

$$\begin{gathered} \frac{S_3}{S_6}=\frac{125}{152} \\ \Rightarrow \frac{{\displaystyle \frac{a\left(r^3-1\right)}{r-1}}}{{\displaystyle \frac{a\left(r^6-1\right)}{r-1}}}=\frac{125}{152} \end{gathered}$$

$\begin{array}{rcl}& \Rightarrow & \frac{\left(r^3-1\right)}{\left(r^6-1\right)}=\frac{125}{152}\\ & \Rightarrow & \frac{\left(r^3-1\right)}{\left(r^3-1\right)\left(r^3+1\right)}=\frac{125}{152}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{]}\\ & \Rightarrow & \left(r^3+1\right)=\frac{152}{125}\\ & \Rightarrow & r^3=\frac{152}{125}-1\\ & \Rightarrow & r^3=\frac{27}{125}\\ & \Rightarrow & r=\frac{3}{5}\end{array}$

Hence, the correct option is D.