Question

If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first ( p + q ) terms.

Answer 1

The sum of first p terms of an A.P. is equal to the sum of the q terms of that A.P.

Let a and d be the first termand common difference of the arithmetic progression respectively.

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

S n = n 2 [ 2a+( n−1 )d ](1)

Let S p , S q be the sum of first p and q terms of the arithmetic progression.

Substitute the values to obtain the sum of first p terms of arithmetic progression in equation (1).

S p = p 2 [ 2a+( p−1 )d ]

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

S q = q 2 [ 2a+( q−1 )d ]

As per the question,

S p = S q (2)

Substitute the values of S p and S q in equation (2).

p 2 [ 2a+( p−1 )d ]= q 2 [ 2a+( q−1 )d ] p[ 2a+( p−1 )d ]=q[ 2a+( q−1 )d ] 2ap+pd( p−1 )=2aq+qd( q−1 ) 2a( p−q )+d[ p( p−1 )−q( q−1 ) ]=0

Further simplify the above expression.

2a( p−q )+d[ p 2 −p− q 2 +q ]=0 2a( p−q )+d[ ( p 2 − q 2 )−( p−q ) ]=0 2a( p−q )+d[ ( p−q )( p+q )−( p−q ) ]=0 2a( p−q )+d[ ( p−q )( p+q−1 ) ]=0

Further simplify the above expression.

( p−q )[ 2a+d( p+q−1 ) ]=0

Equate the terms on both the sides.

p−q≠0     ∴p≠q 2a+d( p+q−1 )=0 d= −2a p+q−1 (3)

Substitute the value of n as p+q in equation (1).

S p+q = p+q 2 [ 2a+( p+q−1 )d ]

Substitute the value of dfrom equation (3) in the above expression.

S p+q = p+q 2 [ 2a+( p+q−1 )× −2a p+q−1 ] = p+q 2 [ 2a−2a ] =0

Thus, the sum of first ( p+q ) terms is 0.