Question

if the sun of first p terms of an AP is same as the sum of its first q terms, then show that sum of its first (p+q) terms is 0.​

Answer 1

Answer:

S  p =S  q

⇒   2 p  (2a+(p−1)d)=  2 q (2a+(q−1)d)

⇒ p(2a+(p−1)d)=q(2a+(q−1)d)

⇒ 2ap+p  2 d−pd=2aq+q  2 d−qd

⇒ 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

⇒ S  p+q  =0

HENCE PROVED

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Answer 2

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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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