Question

Show that the sum of all terms of an A.P. whose first term is a, the second term is b and the let term is c is equal to `((a + c)(b + c - 2a))/(2(b - a))`.

Answer 1

Given, first term, A = a and second term = b

⇒ Common difference, d = b – a

Last term, l = c

⇒ A + (n – 1) d = c

⇒ a + (n – 1) d = c

a + (n – 1)(b – a) = c

⇒ (b – a) (n – 1) = c – a

⇒ `n - 1 = (c - a)/(b - a)`

⇒ `n = (c - a)/(b - a) + 1`

⇒ `n = (c - a + b - a)/(b - a)`

⇒ `n = (b + c - 2a)/(b - a)`

Now, sum = `n/2 [A + l]`

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

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

Hence Proved.