The first, second and middle terms of an AP are a, b, c respectively. Their sum is

First term =a, Second term = b ∴  d=Common difference =b–a It is given that the middle term is c. This means that there are an odd number of tems in the A.P. Let there be (2n+1) terms in the A.P. Then, (n+1)th term is the middle term. ∴   Middle term =c⇒   a+nd  =c ⇒  a+n(b–a)  =  c  ⇒  n  =  c–ab–a ∴    sum  =  2n+12[2a+(2n+1–1)d] ⇒   sum  =  12  {2(c–ab–a)+1}{2a+2(c–ab–a)(b–a)} ⇒    sum  =  12  {2(c–a)b–a+1}{2c}  =  2c(c–a)b–a  +c