a, b and 2a are in A.P so a is the first term and 2a is the last term denoted by T and Tn respectively. Here Common difference = b – a
Tn = 2a = a + (n–1) (b–a)
So n = \(\frac{b}{b-a}\)
Sum = \(\frac{n}{2}\){first term + last term}
= \(\frac{b}{2(b-a)}\{3a\}\)
= \(\frac{3ab}{2(b -a)}\)