Show that the sum of all terms of an A.P. whose first term is $a$, the second term is $b$ and the last term is $c$ is equal to $\frac{(a + c)(b + c - 2a)}{2(b - a)}$.

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Step 1: Here $d = b - a$
Step 2: Let $c$ be the $n^{th}$ term. $c = a + (n - 1)(b - a)$
Step 3: $	herefore n = \frac{c + b - 2a}{b - a}$
Step 4: $\Rightarrow S_n = \frac{c + b - 2a}{2(b - a)} 	imes (a + c)$

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