Question

if the pth term of an AP is 1/q and the qth term is 1/p show that the sum of the pm term is 1/2 (pq+1)

Answer 1

Given pth term = 1/q

a + (p - 1)d = 1/q

aq + (pq - q)d = 1  --- (1)

Similarly, 

 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 2

STEP-BY-STEP EXPLANATION :-

$a + (p - 1)d = \frac{1}{q}.....(i) \\ \\ a + (q - 1)d = \frac{1}{p} .......(ii) \\ \\ \\ \\ subtracting \: them \: \\ \\ \\ a + (p - 1)d - a - (q - 1)d = \frac{p - q}{pq} \\ \\ \\ d(p - 1 - q + 1) = \frac{p - q}{pq} \\ \\ d(p - q) = \frac{p - q}{pq} \\ \\ d = \frac{1}{pq} \\ \\ \\ putting \: in \: (i) \\ \\ a + (p - 1) \frac{1}{pq} = \frac{1}{q} \\ \\ a + \frac{p}{pq} - \frac{1}{pq} = \frac{1}{q} \\ \\ a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q} \\ \\ a = \frac{1}{pq}$

Now,

We have a as well as d.

So,

Sum of first pq terms will be :-

$s = \frac{pq}{2} (2a + (pq - 1)d) \\ \\ = \frac{pq}{2} (2 \times \frac{1}{pq} + (pq - 1) \frac{1}{pq} ) \\ \\ \\ = \frac{pq}{2} ( \frac{2 + pq - 1}{pq} ) \\ \\ \\ \frac{1}{2} (pq + 1)$

Hence,

Proved!