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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)

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Sol.    For an A.P, Let ‘a’ be the first term and ‘d’ be the common difference.
∴                                                                  Sp = Sq                                 [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(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
∴       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.

Arithmetic Progression September 28 , 2018 0 Comments 137 views