Question

1. If the first term of an A.P. is a, the second term
b and the last term c, then show that the sum of
the terms of the A.P. is
(a + c)(b +c-2a)
2 (b-a)
​

Answer 1

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Answer 2

AnswEr:

Let d be the common difference of the given A.P. Then, d = b - a . Let these be n terms in the given A.P. Then,

$\sf \qquad \: c = nth \: term \\ \\ \implies \sf \: c = a + (n - 1)d \\ \\ \implies \sf \: c = a + (n - 1)(b - a) \\ \\ \implies \sf \: n - 1 = \frac{c - a}{b - a} \\ \\ \implies \sf \: n = \frac{c - a}{b - a} + 1 \\ \\ \implies \sf \: n = \frac{b + c- 2a}{b - a}$

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$\therefore \rm \: Sum \: of \: the \: A.P. = Sum \: of \: its \: n \: term \\ \\ \qquad \sf = \frac{n}{2} (a + c) \\ \\ \qquad \sf = \frac{(a + c)(b + c - 2a)}{2(b - a)}$

Hence, Proved !