Question

If the sum of the first p terms of an AP is the same as the sum of its
first a terms (where p#q) then show that the sum of its first (p+q)
terms is zero.
:ac​

Answer 1

p/2[2a +(p-1)d]=q/2 [2a+(q-1)d]

p[2a +pd - d]=q[2a+qd - d ]

2ap + p2d - pd =2aq + q2d -qd

2ap-2aq +p2d -q2d -pd +qd =0

2a (p-q) +(p+q)(p-q)d -d(p-q)=0

(p-q)[2a + (p+q)d - d ]=0

2a + (p+q)d - d=0

2a + [(p+q)-1]d=0

........ .........

Thanks

Mark branliest

@ Run free and dive into the sky.

Answer 2

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

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

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