Question

If the sum of first p terms of an AP is same as the sum of its first q terms (p is not equal to q), then show that the sum of its first (p+q) terms is 0

Answer 1

I hope this will help you

Answer 2

${\green {\boxed {\mathtt {✓verified\:answer}}}}$

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

${\huge{\underline{\underline{\underline{\orange{\mathtt{❢◥ ▬▬▬▬▬▬... ...◆ ......▬▬▬▬▬▬ ◤❢}}}}}}}$