Q.19) If the pth term of an AP be and qth term by -, thenprove that th

1.	Q.19) If the pth term of an AP be and qth term by -, thenprove that the sum of the first pq terms must be (pg+ 1)2
Answer» Given pth term = 1/q That is ap= a + (p - 1)d = 1/qaq + (pq - q)d = 1 --- (1) Similarly, we getap + (pq - p)d = 1 --- (2) From (1) and (2), we getaq + (pq - q)d =ap + (pq - p)daq - ap = d[pq - p - pq + q]a(q - p) = d(q - p) Therefore, a = dEquation (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]

Q.19) If the pth term of an AP be and qth term by -, thenprove that the sum of the first pq terms must be (pg+ 1)2

Answer»

Given pth term = 1/q

That is ap= a + (p - 1)d = 1/qaq + (pq - q)d = 1 --- (1)

Similarly, we getap + (pq - p)d = 1 --- (2)

From (1) and (2), we getaq + (pq - q)d =ap + (pq - p)daq - ap = d[pq - p - pq + q]a(q - p) = d(q - p)

Therefore, a = dEquation (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]