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

Solution:

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

Let the A.P. be a, a+d, a+2d........a+nd.

Here, let d be the common difference and n be the total number of terms.

Given:

$a_{1}=a$

$a_{2}=b$

$\Rightarrow a+d=b$

$\Rightarrow d=b-a$    ....(1)

$a_{n}=2 a$

$\Rightarrow a+(n-1) d=2 a$

$\Rightarrow(n-1) d=a$

$\Rightarrow d=\frac{a}{n-1} \quad \ldots . .(2)$

From equations $(1)$ and $(2)$, we have:

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

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

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

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

Now, sum of n terms of an A.P.:

$S=\frac{n}{2}\left\{a+a_{n}\right\}$

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

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