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If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

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Solution: we have given


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

Multiply both sides with (p + q)/2

by (12 points)
2a + [(p+q)-1]d=0
How did (1) came?
by (29.2k points)
@Shabbir  2a + (p+q)d - d=0 taking d as common from the equation we will get

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

Everything is elaborated in the question, I hope it helps you

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