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

In the given problem, we are given first, second and last term of an A.P. We need to find its sum.

So, here

First term = a

Second term (a2) = b

Last term (l) = 2a

Now, using the formula $a_{n}=a+(n-1) d$

$a_{2}=a+(2-1) d$

$b=a+d$

$d=b-a$ ...........(1)

Also,

$a_{n}=a+(n-1) d$

$2 a=a+n d-d$

$2 a-a=n d-d$

$\frac{a+d}{d}=n$ ...........(2)

Further as we know,

$S_{n}=\frac{n}{2}[a+l]$

Substituting (2) in the above equation, we get

Using (1), we get

$S_{n}=\frac{a+(b-a)}{2(b-a)}(3 a)$

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

Thus,

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

Therefore, the correct option is (c).