Math, asked by shrigurukrupaksk1234, 1 year ago

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)

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ananthram003: good question.

Answers

Answered by skh2
92

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!


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ritika200119: nice ans
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Answered by UltimateMasTerMind
100

<b>

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.


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