tn an A.P, if ph rmiand5. Interm is prove that the sum of firitof firs

1.	tn an A.P, if ph rmiand5. Interm is prove that the sum of firitof first pas is 1(m +1), where p # q.terms iS
Answer» 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]

tn an A.P, if ph rmiand5. Interm is prove that the sum of firitof first pas is 1(m +1), where p # q.terms iS

Answer»

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]