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

Answer: Option C

First term (a₁) = a
Second term (a₂) = b
and last term (l) = 2a
∴ d = Second term - First term = b - a
∴ l = a_n = a + (n - 1)d
$$\begin{array}{rl}& \Rightarrow 2a=a+\left(n-1\right)\left(b-a\right)\\ & \Rightarrow \left(n-1\right)\left(b-a\right)=a\\ & \Rightarrow n-1=\frac{a}{b-a}\\ & \Rightarrow n=\frac{a}{b-a}+1\\ & \Rightarrow n=\frac{a+b-a}{b-a}\\ & \Rightarrow n=\frac{b}{b-a}\\ & \therefore S_n=\frac{n}{2}\left[a+l\right]\\ & =\frac{b}{2\left(b-a\right)}\left[a+2a\right]\\ & =\frac{3ab}{2\left(b-a\right)}\end{array}$$