Question

If the first, second and last term of an A.P. are$a,b$ and $2a$ respectively. then its sum is

• A. $\frac{ab}{\left(b-a\right)}$
• B. $\frac{ab}{2\left(b-a\right)}$
• C. $\frac{3ab}{2\left(b-a\right)}$
• D. $\frac{3ab}{4\left(b-a\right)}$

Answer 1

The correct option is C

$\frac{3ab}{2\left(b-a\right)}$

Explanation for the correct option

Step 1: Solve for the number of terms in the given progression

Given that, first, second and last term of an A.P. are$a,b$ and $2a$ respectively

Let $d$ be the common difference between two successive terms of the arithmetic progression

$a,b$ are successive terms in the A.P then $d=b-a$

The $n^{th}$ term of the A.P is given as $a_n=a+\left(n-1\right)d$ where $a$ is the first term and $n$ is the number of terms in the A.P

Consider the $n^{th}$ term to be the last term of the A.P. Substituting all the required values we get,

$2a=a+\left(n-1\right)\left(b-a\right)$

$\Rightarrow$ $\frac{a}{b-a}=\left(n-1\right)$ $...\left[1\right]$

$\Rightarrow$$\frac{a}{b-a}+1=n$

$\Rightarrow$$\frac{a+b-a}{b-a}=n$

$\Rightarrow$ $\frac{b}{b-a}=n$

Step 2: Solve for the sum

The sum of $n$ terms of an A.P is given as $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

Substituting all the required values we get

$S_n=\frac{b}{2\left(b-a\right)}\left[2a+\frac{a}{b-a}\times \left(b-a\right)\right]$ [from $\left[1\right]$]

$\Rightarrow S_n=\frac{3ab}{2\left(b-a\right)}$

Thus the sum of all the terms of the given A.P is $\frac{3ab}{2\left(b-a\right)}$.

Hence option(C) i.e. $\frac{3ab}{2\left(b-a\right)}$ is the correct answer.