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

True

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False

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Solution

The correct option is A True
$T_{1} = a, T_{2} = a + d = b$
$∴ d = b - a$
$T_{n} = a + (n - 1) d = c$
$n - 1 = \frac{c - a}{d} = \frac{c - a}{b - a}$
$∴ n = \frac{c - a}{b - a} + 1 = \frac{c + b - 2 a}{b - a}$ .. $(1)$
$S_{n} = \frac{n}{2} [a + T_{n}] = \frac{(c + b - 2 a)}{2 (b - a)} (a + c)$

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Similar questions

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)

a

b

c

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

a

b

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

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

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

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