# Arithmetic progression 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 where pis not equal to q

Let a be the first term and common difference is d.
Sum of first n terms is S(n) = (n/2) { 2a + (n-1) d}

Given S(p) = S(q)
=> (p/2) [ 2a + (p-1)d ] = (q/2) [ 2a + (q-1)d ]

=>  2ap + (p2-p)d = 2aq + (q2-q)d
=>  2ap + (p2-p)d - 2aq - (q2-q)d =0

=> 2a(p-q) + (p2-p-q2+q)d =0
=> 2a(p-q) + (p2-q2)d - (p-q)d = 0
=> 2a(p-q) + (p-q)[ (p+q) - 1]d = 0
=> (p-q) [ 2a + (p+q-1)d ] = 0
=> 2a + ( p+q-1)d = 0  {as p and q are not equal}

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

Hence proved.

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