If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)
We know that, sum of nth term of an A.P. we will find it’s
sum
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Where, n = no. of terms
a = first term
d = common difference
Sn = sum of n terms
Now, Sum of p terms is
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And, Sum of q terms is
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Given: Sp = Sq
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Multiply by 2 on both sides, we get,
⇒p[ 2a + (p – 1)d] = q[ 2a + (q – 1)d]
⇒2ap + p(p – 1)d = 2aq + q(q – 1)d
⇒ 2ap – 2aq + p(p – 1)d – q(q – 1)d = 0
⇒ 2a(p – q) + d[p2 – p– q2 + q] = 0
⇒ 2a(p – q) + d[(p2– q2 ) – p + q] = 0
⇒ 2a(p – q) + d[(p– q )(p + q) – (p – q)] = 0
(since, (a – b)(a + b) = a2 – b2)
⇒ 2a(p – q) + d(p – q) [p + q – 1 ] = 0
⇒ (p – q)[2a + d (p + q – 1) ] = 0
Since, p ≠ q
∴ p – q ≠ 0
⇒ 2a + d (p + q – 1) = 0
Multiply both side by ![]()
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⇒ Sp + q = 0
Hence proved
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