If the first ,second and the last term of an A.P are $a, b, c$ respectively,the the sum is ?

\frac{(a + b) (a + c - 2 b)}{2 (b - a)}

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\frac{(b + c) (a + b - 2 c)}{2 (b - a)}

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\frac{(a + c) (b + c - 2 a)}{2 (b - a)}

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None of these

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Solution

The correct option is D $\frac{(a + c) (b + c - 2 a)}{2 (b - a)}$
$∵ a_{1} = a a_{2} = b a_{n} = c h e n c e, d = a_{2} - a_{1} = b - a ∵ a_{n} = c \Rightarrow a_{1} + (n - 1) d = c \Rightarrow a + (n - 1) (b - a) = c \Rightarrow n = \frac{(b + c - 2 a)}{(b - a)} ∴ S_{n} = \frac{n}{2} [a + a_{n}] = \frac{(b + c - 2 a)}{2 (b - a)} [a + c)] = \frac{(a + c) (b + c - 2 a)}{2 (b - a)}$

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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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