Question

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

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

Answer 1

Answer: (C)

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

Common difference $=b-a$

As we know the nth term, $a_n=a+(n-1)d$, where a and d are the first term and common difference respectively.

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

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

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

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

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

$\therefore$ Sum of $AP=\frac{n}{2}[a+2a]$

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