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Question Number 41966 by Raj Singh last updated on 16/Aug/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
	$S_p =(p/2)[2a+(p−1)d] S_Q =(Q/2)[2a+(Q−1)d] given S_p =S_Q (p/2)[2a+(p−1)d]=(Q/2)[2a+(Q−1)d] 2pa+(p^2 −p)d−2Qa−(Q^2 −Q)d=0 2a(P−Q)+d(P^2 −P−Q^2 +Q)=0 2a(P−Q)+d{(P+Q)(P−Q)−(P−Q)}=0 2a(P−Q)+(P−Q){P+Q−1}d=0 2a+(P+Q−1)d=0 nos S_(P+Q) =((P+Q)/2)[2a+(P+Q−1)d] =((P+Q)/2)×0=0$
	$S_{p} = \frac{p}{2} [2 a + (p - 1) d]$ $S_{Q} = \frac{Q}{2} [2 a + (Q - 1) d]$ $g i v e n S_{p} = S_{Q}$ $\frac{p}{2} [2 a + (p - 1) d] = \frac{Q}{2} [2 a + (Q - 1) d]$ $2 p a + (p^{2} - p) d - 2 Q a - (Q^{2} - Q) d = 0$ $2 a (P - Q) + d (P^{2} - P - Q^{2} + Q) = 0$ $2 a (P - Q) + d {(P + Q) (P - Q) - (P - Q)} = 0$ $2 a (P - Q) + (P - Q) {P + Q - 1} d = 0$ $2 a + (P + Q - 1) d = 0$ $n o s S_{P + Q} = \frac{P + Q}{2} [2 a + (P + Q - 1) d]$ $= \frac{P + Q}{2} \times 0 = 0$
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