Question 7
Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to $\frac{(a + c) (b + c - 2 a)}{2 (b - a)}$

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Solution

Given that, the AP is a, b, …..c
Here, first term = a, common difference = b – a
And last term, $l = a_{n} = c$
$∵ a_{n} = l = a (n - 1) d$
$\Rightarrow c = a + (n - 1) (b - a)$
$\Rightarrow (n - 1) = \frac{c - a}{b - a}$
$\Rightarrow n = \frac{c - a}{b - a} + 1$
$\Rightarrow n = \frac{c - a + b - a}{b - a} = \frac{c + b - 2 a}{b - a}$
$∴ S u m o f a n A P, S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$= \frac{(b + c - 2 a)}{2 (b - a)} [2 a + {\frac{b + c - 2 a}{b - a} - 1} (b - a)]$
$= \frac{(b + c - 2 a)}{2 (b - a)} [2 a + \frac{c - a}{b - a} . (b - a)]$
$= \frac{(b + c - 2 a)}{2 (b - a)} (2 a + c - a)$
$= \frac{(b + c - 2 a)}{2 (b - a)} . (a + c)$

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Show that the sum of a A.P. whose first term is 'a', the second term is 'b' and the last term is 'c', is equal to (a+c)(b+c-2a)/2(b-a)

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