Question

IF pth term of an A.P. is 1/q and qth term is 1/p then prove that the sum of the first pq term is 1/2[pa+1]

Answer 1

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]

Answer 2

Question shd be "If pth term of an A.P is 1/q and its qth term is 1/p then show that its sum of (pq)th terms is ---- 1/2 (pq + 1)."

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]