Question

If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is  \[\frac{\left( a + c \right) \left( b + c - 2a \right)}{2\left( b - a \right)}\].

Answer 1

a, b, ..., c

t₁ = a, d = b - a, t_n = c

We know that

t_n = a + (n - 1)d

c = a + (n - 1)(b - a)

`(c-a)/(b-a)=n-1` 

`(c-a)/(b-a)+1/1=n`

`(c-a+b-a)/(b-a)=n`

∴ n = `(c+b-2a)/(b-a)`      ...(1)

Now, 

S_n = `n/2[t_1 + t_n]`

S_n = `(c+b-2a)/((b-a)2)[a+c]`

S_n = `((a+c)(b+c-2a))/(2(b-a))`

Hence proved.