Question

In an A.P. if $p^{th}$ term is $\frac{1}{q}$ and the $q^{th}$ term is $\frac{1}{p}$, prove that the sum of first $pq$ terms is $\frac{1}{2}(1+pq)$, where $p \neq q$.

Answer 1

Answer:

Sum of pq is $\dfrac{1}{2}$( 1 + pq )

Step-by-step explanation:

Let a be the first term of the AP and d be the common difference of the AP, then

$a_{p}=\dfrac{1}{q}\:\:and\:\:a_{q} = \dfrac{1}{p}$

Now,

= >  pth term = 1 / q

= >  a + ( p - 1 )d = 1 / q    ...( i )

And,

= >  qth term = 1 / p

= >  a + ( q - 1 )d = 1 / p  ...( ii )

Now, subtract ( ii ) from ( i ),

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

= >  dp - d - qd + d = ( p - q ) / pq

= >  dp - dq = ( p - q ) / pq

= >  d( p - q ) = ( p - q ) / pq

= >  d = 1 / pq

Substitute the value of d in ( i ),

= >  a  + ( p - 1 )d = 1 / q

= >  a + ( p - 1 )( 1 / pq ) = 1 / q

= >  a + ( p / pq )  - ( 1 / pq ) = 1 / q

= >  a + 1 / q - 1 / pq = 1 / q

= >  a = 1 / pq

Sum of pq terms  :

= >  ( pq / 2 ) x [ 2a + ( pq  - 1 )d ]

Substitute the value of a and d : -  

= >  pq / 2 x [ 2( 1 / pq ) + ( pq - 1 )( 1 / pq ) ]

= >  pq / 2 x [ ( 2 / pq ) + ( pq - 1 )( 1 / pq ) ]

= >  pq / 2pq x [ 2 + pq - 1 ]

= >  1 / 2 x [ 2 + pq - 1 ]

= >  1 / 2 x [ 1 + pq ]

Hence proved that the sum of pq terms is $\dfrac{1}{2}(1+pq)$.

Answer 2

It will help you dude....