Question
Fri November 11, 2011 By:

# If in an AP the sum of p terms is equal to q and the sum of q terms is equal to p, then prove that the sum of (p+q) terms is -(p+q)

Fri November 11, 2011
Sum of n terms for an AP = n[2a+(n-1)d]/2

Sp = q

Put n = p in the formula above to get

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

Or,  2q/p = 2a + (p-1)d   ......................(1)

And Sq = p

put n=q we get

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

Or, 2p/q = 2a+ (q-1)d   ...............................(2)

Subtract eqn 2 from 1 we get

2q/p - 2p/q = (p-q)d

2(q2-p2)/pq = (p-q)d

2(q-p)(q+p)/pq= (p-q)d

d= -2(p+q)/pq

Put this value of d in eqn 2 we get

2p/q = 2a - 2(q-1)(p+q)/pq

a = p/q + (q-1)(p+q)/pq

a = p2 + pq + q2 - p - q / pq

Therefore
Sp+q = (p+q)[2a + (p+q-1)d]/2

= (p+q)[2(p2 + pq + q2 - p - q)/pq -2 (p+q-1)(p+q)/pq]/2

= (p+q)[(p2 + pq + q2 - p - q)- (p+q-1)(p+q)]/pq

=(p+q)[(p2 + pq + q2 - p - q)- (p2+pq+qp+q2-p-q]/pq

= (p+q)[p2 + pq + q2 - p - q- p2-pq-qp-q2++p+q]/pq

= (p+q)[-pq]/pq

=-(p+q)

Hence Proved

Thank you
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