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

Let first term of A.P be a

Second term of A.P be b and last term of A.P be given by 2a

Since d = a2 – a1 = b – a

$n^{\text {th }}$ term i. e $a_{n}=a+(n-1) d$

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

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

i. e $n-1=\frac{a}{b-a}$ i. e $n=\frac{a}{b-a}+1=\frac{b}{b-a}$

∴ Sn i.e sum of n terms of A.P is

$\frac{n}{2}\left(a_{1}+a_{n}\right)$

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

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

$\therefore$ Sum of its term $=\frac{3 a b}{2(b-a)}$