In an A.P., if pth term is 1/q and qth term is 1/p , prove that the sum of first pq terms is (pq +1)/2, where p ≠ q.
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Solution:
Given pth term = 1/q
That is ap = a + (p - 1)d = 1/q
aq + (pq - q)d = 1 --- (1)
Similarly, we get ap + (pq - p)d = 1 --- (2)
From (1) and (2), we get
aq + (pq - q)d = ap + (pq - p)d
aq - ap = d[pq - p - pq + q]
a(q - p) = d(q - p)
Therefore, a = d
Equation (1) becomes,
dq + pqd - dq = 1
d = 1/pq
Hence a = 1/pq
Consider, Spq = (pq/2)[2a + (pq - 1)d]
= (pq/2)[2(1/pq) + (pq - 1)(1/pq)]
= (1/2)[2 + pq - 1]
= (1/2)[pq + 1]
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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
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