Question

If the sum of the first p terms of an AP is the same as the sum of its
first q terms (where p #q) then show that the sum of its first (p+q)
terms is zero.
Let a be the first term and d be the common difference of the
given AP. Then,
9
S = S, = [22+(p. – 1)d]
[2a + (p-1)d] = 3 [2a + (9-1)d]
> (p - 9) (2a) = (9-p) (q+p-1)d
= 2a = (1-p-9)d
Sum of the first (p+q) terms of the given AP
(p+q)
{2a + (p+q-1)d
... (i)
2




please answer me soon​

Answer 1

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$let \: \: a \: \: be \: the \: first \: term \: and \: d \: be \: the \: common \: difference \: of \: the \: given \: ap \: \\ then \\ s _{p} = s _{q} \implies \frac{p}{2} (2a + (p - 1)d) = \frac{q}{2} (2a + (q - 1)d \\ \implies(p - q)(2a) = (q - p)(q + p - 1) \\ \implies2a = (1 - p - q)d \: \: \: \: \: \: \: .....(1) \\ sum \: of \: the \: first \: (p + q) \: terms \: of \: the \: given \: ap \\ = \frac{(p + q)}{2} (2a + (p + q - 1)d) \\ = \frac{(p + q)}{2} .(1 - p - q)d + (p + q - 1)d \: \: \: \: \: \: \: \: (using \: 1) \\ = 0$

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