The first , second and the last terms of an A.P. are `a ,b , c`respectively. Prove that the sum is `((a+c)(b+c)(c-2a))/(2(b-a))`.

The first , second and the last terms of an A.P. are `a ,b , c`respect

1.	The first , second and the last terms of an A.P. are `a ,b , c`respectively. Prove that the sum is `((a+c)(b+c)(c-2a))/(2(b-a))`.
Answer» Let d be the common difference of the given AP and let it contain n terms. Then d=(b -a) Now, ` T_(n) = c Rightarrow a + (n-1) d=c` ` Rightarrow a+ (n-1) (b-a) =c` ` Rightarrow (n-1) = ((c-a))/((b-a))` `Rightarrown= { ((c-a))/((b-a))+1}=((b+c-2a))/((b-a))` sum of the given `AP= n/2 (a+l)` , where l is the last term `((b+c-2a))/(2(b-a))xx (a+c)` ` [ because n= ((b+c-2a))/(2(b-a))andl=c]`j `((a+c) (b+c-2a))/(2(b-a))`