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 \(\frac{(a + c)(b + c - 2a)}{2(b - a)}\).
Answer: Proved, \(S = \frac{(a + c)(b + c - 2a)}{2(b - a)}\)
Common difference \(d = b - a\). Let there be \(n\) terms; the last term is \(c\):
\[c = a + (n - 1)(b - a)\]\[n - 1 = \frac{c - a}{b - a}\]\[n = \frac{c - a + b - a}{b - a} = \frac{b + c - 2a}{b - a}\]\[S = \frac{n}{2}(\text{first} + \text{last})\]\[= \frac{b + c - 2a}{2(b - a)}(a + c)\]Hence \(S = \frac{(a + c)(b + c - 2a)}{2(b - a)}\).
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