Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to `((a+c)(b+c-2a))/(2(b-a))`.

Explanation

Given that, the AP is a, b, …, c.
Here, first term = a, common difference = b - a
and last term ,`l=a_(n)=c`
` :' a_(n)=l=a+(n-1)d`
`implies c=a+(n-1)(b-a)`
`implies (n-1)=(c-a)/(b-a)`
`implies n=(c-a)/(b-a)=1`
` implies n=(c-a+b-a)/(b-a)=(c+b-2a)/(b-a) " " ` ...(i)
` :. ` Sum of an AP, ` S_(n)=(n)/(2)[2a+(n-1)d]`
`" " =((b+c-2a))/(2(b-a))[2a+{(b+c-2a)/(b-a)-1}(b-a)]`
` =((b+c-2a))/(2(b-a))[2a+(c-a)/(b-a).(b-a)]`
`=((b+c-2a))/(2(b-a))(2a+c-a)`
` =((b+c-2a))/(2(b-a)).(a+c) " " ` Hence proved.