If sum of 1st p terms of an AP is equal to the sum of 1st q terms ,then show that the sum of its 1st (p+q) terms is zero,where p is not equal to q.
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Dear Student,
If a is the first term and d the common difference of an AP, then
S(n) = (n/2) [ 2a + (n-1)d.
Given : S(p) = S(q), ... [ where ... p ≠ q.]
∴ (p/2) [ 2a + (p-1)d ] = (q/2) [ 2a + (q-1)d ]
∴ 2ap + p(p-1)d = 2aq + q(q-1)d
∴ ( 2ap - 2aq ) + [ (p²-p) - (q²-q) ]d = 0
∴ 2a(p-q) + [ (p²-q²) - (p-q) ]d = 0
∴ 2a(p-q) + (p-q)[ (p+q) - 1]d = 0
∴ dividing throughout by (p-q),
. . 2a + ( p+q-1)d = 0
which is the sum of first (p+q) terms of the AP.
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