Question

Sum of the first p, q and r terms of an A.P. are a, b and c , respectively. Prove that

Answer 1

The sum of p, q, r terms of an A.P. are a, b, c respectively.

Let a 1 and d be the first term and common difference of the A.P. respectively.

The formula for the sum of n terms in A.P. is given by,

S n = n 2 [ 2a+( n−1 )d ](1)

Substitute the value in equation (1) to obtain the sum of p terms.

S p = p 2 [ 2 a 1 +( p−1 )d ] a= p 2 [ 2 a 1 +pd−d ] 2a p =[ 2 a 1 +pd−d ] (2)

Similarly, substitute the values in equation (1) to obtain the sum of q terms.

S q = q 2 [ 2 a 1 +( q−1 )d ] b= q 2 [ 2 a 1 +qd−d ] 2b q =[ 2 a 1 +qd−d ] (3)

Similarly, substitute the values in equation (1) to obtain the sum of r terms.

S r = r 2 [ 2 a 1 +( r−1 )d ] c= r 2 [ 2 a 1 +rd−d ] 2c r =[ 2 a 1 +rd−d ] (4)

Subtract the equation (3) from equation (2).

( p−1 )d−( q−1 )d= 2a p − 2b q d( p−1−q+1 )= 2aq−2bp pq d( p−q )= 2aq−2bp pq d= 2aq−2bp pq( p−q ) (5)

Subtract the equation (4) from equation (3).

( q−1 )d−( r−1 )d−= 2b q − 2c r d( q−1−r+1 )= 2br−2cq qr d( q−r )= 2br−2cq qr d= 2br−2cq qr( q−r ) (6)

Equate the values of d from equation (5) and equation (6).

2aq−2bp pq( p−q ) = 2br−2cq qr( q−r ) 2( aq−bp ) pq( p−q ) = 2( br−cq ) qr( q−r ) aq−bp pq( p−q ) = br−cq qr( q−r ) qr( q−r )( aq−bp )=pq( p−q )( br−cq )

Further simplify the above expression.

r( q−r )( aq−bp )=p( p−q )( br−cq ) ( aqr−bpr )( q−r )=( pbr−pqc )( p−q )

Divide the above expression by pqr on both the sides.

( aqr−bpr ) pqr ( q−r )= ( pbr−pqc ) pqr ( p−q ) ( a p − b q )( q−r )=( b q − c r )( p−q ) a p ( q−r )− b q ( q−r )= b q ( p−q )− c r ( p−q ) a p ( q−r )− b q ( q−r )− b q ( p−q )+ c r ( p−q )=0

Further simplify the above expression.

a p ( q−r )− b q ( q−r+p−q )+ c r ( p−q )=0 a p ( q−r )− b q ( p−r )+ c r ( p−q )=0 a p ( q−r )+ b q ( r−p )+ c r ( p−q )=0

Hence, the given statement is proved.