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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)

Print or SaveSol. For an A.P, Let ‘a’ be the first term and ‘d’ be the common difference.

∴ S_{p} = S_{q} [Given]

We know that for an A.P, Sn = \({n\over2}[2a+(n-1)d]\)

∴ \({p\over2}[2a+ ( p– 1)d]\) = \({q\over2}[2a+(q-1)d]\)

Multiplying by 2 on both the 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[p(p – 1) – q(q – 1)] = 0

∴ 2a(p – q) + d(p^{2} – p – q^{2} + q) = 0

∴ 2a(p – q) + d(p^{2} – q^{2} – p + q) = 0

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

∴ 2a(p – q) + d(p + q) (p – q) – d(p – q) = 0

p ≠ q [Given]

∴ p – q ≠ 0

Dividing throughout by (p – q), we get;

2a + d(p + q) – d = 0 ...(I)

2a + d(p + q – 1) = 0

We have to prove S(p + q) = 0

∴ S(p + q) = \({{p+q}\over2}[2a+(p+q-1)d]\)

∴ S(p + q) = \({{p+q}\over 2}*(0)\) [From (I)]

∴ S(p + q) = 0

∴ ** The sum of the first p + q terms is zero.**