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If the pth term of an H.P. is q the qth term is p ,

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prove that the (p+q)th is pq/p+q

asked Jun 14, 2013 in ALGEBRA 2 by futai Scholar

1 Answer

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As the relation between HP & AP, is the reciprocal of each term of HP forms AP,

pth term in AP is 1/q  and that qth term in AP is 1/p

In A.P  series  the n^th  term  is  a + ( n - 1) d.

Let the first term of the AP be 'a' and its common difference be 'd'

So pth term of AP: a + (p - 1)d = 1/q ------ (1)

and qth term of AP is: a + (q - 1)d = 1/p ------ (2)

Subtract  equation (1) and  (2)

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

d(p - q)= (p - q)/pq

DIvide each side by  ( p - q )

Therefore d = 1/pq

Substitute this value of d in any one of the above two equations

a +( q - 1 ) 1/pq  =  1/p

( q - 1 )/ pq  =  1/p -a

q/pq - 1/ pq  = 1/p  -a

Therefore  a = 1/ pq

So ( p+q )th  term in  A .P  a + ( p+ q -1 ) d

 =  ( 1/ pq ) + ( p+ q -1) / 1/pq

 = ( p + q ) / pq

Hence  the  term in H.P  is  pq /(p+q).






 

answered Jun 14, 2013 by goushi Pupil

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