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# if pth term of an ap is 1/q and the qth term of an ap is 1/p, show that the sum of pq terms is pq+1/2

Asked by 23rd February 2013, 7:25 PM
Answer : Given :if pth term of an ap is 1/q and the qth term of an ap is 1/p
To prove : the sum of pq terms is pq+1/2

Let first term and common difference be a and d respectively
Now according to the question
Tp=a+(p-1)d=1/q ..................eq (1)
Tq=a+(q-1)d=1/p ...................eq(2)

Subtracting eq 2 by eq1, we get
=> (p-q)d = (1/q )- (1/p )
=> (p-q)d = [ (p-q) / pq ]
=> d =1/pq......................eq 3

Substituting the value of d in eq 1
=> 1/q=a+[(p-1) (1/(pq)) ]
=> 1/q = a + [1/q] - [1/(pq)]
=> a=1/pq.................................eq 4

Now the sum of pq terms using the formula Sn = n/2 ( 2a +(n-1) d ) is given by
=> Spq=(pq/2 ) [ (2/(pq)) + (pq-1) (1/pq)]   {using the value of eq 3 and eq 4}
=> Spq= (pq/2 ) [ (2/(pq)) + 1 -(1/pq)]
=> Spq= (pq/2 ) [ (1/(pq)) + 1]
=> Spq = ( 1/2 ) + (pq /2)
=> Spq=(pq+1)/2
Hence proved
Answered by Expert 23rd February 2013, 9:38 PM
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