Question

In a G. P. the first term is $a$, the second term is $b$ and the last term is $c$, then the sum of the series is.

• A. $\begin{array}{rcl}& & \frac{c^2-ab}{c-a}\end{array}$
• B. $\begin{array}{rcl}& & \frac{b^2-ac}{b-a}\end{array}$
• C. $\begin{array}{rcl}& & \frac{a^2-bc}{a-b}\end{array}$
• D. $\begin{array}{rcl}& & \frac{a^2-bc}{a+b}\end{array}$

Answer 1

The correct option is C

$\begin{array}{rcl}& & \frac{a^2-bc}{a-b}\end{array}$

Explanation for the correct answer:

To find the sum of the series is:

Given,

The first term is $a$

The second term is $b$

The last term is $c$

Let us consider there are $n$ terms in the G.P,

We know the formula for $n^{th}$ term in the G.P is ${\mathbf{t}}_{\mathbf{n}}\mathbf{=}{\mathbf{ar}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$ where $a$ is the first term and $r$ is common ratio.

So,

$\begin{array}{rcl}\frac{b}{a}& =& r,\cdots \cdots \cdots \left(1\right)\\ c& =& a{r}^{n-1}\\ cr& =& a{r}^n\cdots \cdots \cdots \left(2\right)\end{array}$

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

So,

$\begin{array}{rcl}{S}_n& =& \frac{a\left(1-r^n\right)}{1-r}\\ & =& \frac{a-a{r}^n}{1-r}\\ & =& \frac{a-cr}{1-r}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{}\mathbf{from}\mathbf{}\mathbf{equation}\mathbf{}\mathbf{2}\mathbf{]}\\ & =& \frac{a-c\left({\displaystyle \frac{b}{a}}\right)}{1-\frac{b}{a}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{}\mathbf{from}\mathbf{}\mathbf{equation}\mathbf{}\mathbf{1}\mathbf{]}\mathbf{}\mathbf{}\\ & =& \frac{a^2-bc}{a-b}\end{array}$

Hence, the correct option is C.