Question

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

• A. $\frac{ab}{2(b-a)}$
• B. $\frac{3ab}{2(b-a)}$
• C. $\frac{ab}{(b-a)}$
• D. None of these

Answer 1

The correct option is B

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

Explanation for the correct option:

Step1: Solve for number of terms in the series

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

first term $=a$

second term$=b$

common difference $d=$ second term $-$ first term $=$$b-a$

Last term $l=2a$

$\begin{array}{crcl}\Rightarrow & a+\left(n-1\right)d& =& 2a\\ \Rightarrow & \left(n-1\right)\left(b-a\right)& =& a\\ \Rightarrow & n\left(b-a\right)& =& a+\left(b-a\right)\\ \Rightarrow & n& =& \frac{b}{\left(b-a\right)}\end{array}$

Therefore, number of terms $n=\frac{b}{b-a}$

Step 2: Solve for the sum of the series

Sum of the series in Arithmetic progression $S=\frac{a+l}{2}$

$\begin{array}{cccl}& & =& \frac{b}{2\left(b-a\right)}\left(a+2a\right)\\ & S& =& \frac{3ab}{2\left(b-a\right)}\end{array}$

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