Question

Let the sum of n , 2 n , 3 n terms of an A.P. be S 1 , S 2 and S 3 , respectively, show that S 3 = 3 (S 2 – S 1 )

Answer 1

Let, a be the first term and b be the common difference of the A.P.

It is given that the sum of n, 2n and 3n terms of an A.P is S 1 , S 2 and S 3 respectively, so,

S 1 = n 2 [ 2a+( n−1 )d ] S 2 = 2n 2 [ 2a+( 2n−1 )d ] =n[ 2a+( 2n−1 )d ] S 3 = 3n 2 [ 2a+( 3n−1 )d ]

Subtract S 1 from S 2 ,

S 2 − S 1 =n[ 2a+( 2n−1 )d ]− n 2 [ 2a+( n−1 )d ] =n{ 4a+4nd−2d−2a−nd+d 2 } =n{ 2a+3nd−d 2 } = n 2 [ 2a+( 3n−1 )d ]

This shows that,

3( S 2 − S 1 )= 3n 2 [ 2a+( 3n−1 )d ]

Hence, it is proved that S 3 =3( S 2 − S 1 ).