Question

If pth term of AP is 1/q and qth term of A.p is 1/p then prove that sum of first pq terms of AP is 1/2(pq+1)

Answer 1

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!

Answer 2

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Solution:-

Given :-

A p = 1/q

$= > \frac{1}{q} = a + (p - 1)d..........(1)$

&

$= > \frac{1}{p} = a + (q - 1)d..........(2)$

Subtracting eq (1) From (2). we get,

$= > { \: a + (p - 1)d \: } - {a - (q - 1)d} = \frac{1}{q} - \frac{1}{p} \\ \\ = > (p - q)d = \frac{p - q}{pq} \\ \\ = > d = \frac{1}{pq} ....................(3)$

Putting eq 3 in eq (1). we get,

$= > a + (p - 1)d = \frac{1}{q} \\ \\ = > a + (p - 1) \frac{1}{pq} = \frac{1}{q} \\ \\ = > a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q} \\ \\ = > a = \frac{1}{pq}$

Now,

Sum of pq term :-

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

Hence Proved.