If $a, b, c$ be the $1^{s t}, 3^{r d}$ and $n^{t h}$ terms respectively of an $A . P .$ , prove that the sum to $n$ terms is $\frac{c + a}{2} + \frac{c^{2} - a^{2}}{b - a}$ .

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Solution

$a_{1} = a$ & $a_{3} = b$ $= a + 2 d d = \frac{b - a}{2}$
$c = a + (n - 1) d$
$c = a + (n - 1) (\frac{b - a}{2})$
$\frac{2 (c - a)}{b - a} = n - 1$
$\frac{2 (c - a) + b - a}{b - a} = n$
$S_{n} = \frac{n}{2} [2 a + (n - 1) d] S_{n} = \frac{2 (c - a) + b - a}{2 (b - a)} [2 a + \frac{2 (c - a)}{(b - a)} \times \frac{(b - a)}{2}] = \frac{n}{2} [2 a + (n - 1) d] = \frac{2 (c - a) + b - a}{2 (b - a)} (a + c)$
$S_{n} = \frac{2 (c - a) (c + a)}{2 (b - a)} + \frac{(b - a) (c + a)}{2 (b - a)} = \frac{c^{2} - a^{2}}{b - a} + \frac{c + a}{2}$ $(p r o v e d)$

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