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

Here, `t_(1)=a, t_(2)=band t_(n)=c`
the common difference `=d=(b-a)`
Let us find the total number of terms in the A.P.
`t_(n)=a+(n-1)d` …(Formula)
`thereforec=a+(n-1)xx(b-a)`
`thereforec-a=(n-1)xx(b-a)1`
`thereforen-1=(c-a)/(b-a)`
`thereforen=(c-a)/(b-a)+1`
`thereforen=(c-a+b-a)/(b-a)`
`thereforen=(c+b-2a)/(b-a)` i.e.`n=(b+c-2a)/(b-a)`
The sum of `n` term of the A.P.
`S_(n)=(n)/(2)[t_(1)+t_(n)]` ...(Formula)
`=((b+c-2a)/(b-a))/(2)[a+c]` ...(Substituting the values)
`=(b+c-2a)/(2(b-a))(a+c)`
i.e. `S_(n)=((a+c)(b+c-2a))/(2(b-a))`