Question

the sum of n, 2n, 3n terms of an AP are S1, S2 & S3 respectively. Prove that S3 = 3(S2 - S1) ​

Answer 1

Step-by-step explanation:

Sn = n/2 [ 2a + (n-1)d ]

S1 = n/2 [ 2a + (n-1)d ]

S2 = 2n/2 [ 2a + (n-1)d ]

= n [2a + (n-1)d ]

S3 = 3n/2 [ 2a + (n-1)d ]

TO PROVE : S3 = 3(S2 - S1)

S3 = 3(S2 - S1)

S3 = 3S2 - 3S1

3n/2 [ 2a + (n-1)d ] = 3{ n [2a + (n-1)d] }

= -3{ n/2 [ 2a + (n-1)d }

Dividing [ 2a + (n-1)d ] in each terms ,

3n/2 = 3(n) - 3(n/2)

3n/2 = 3n - 3n/2

= (6n/2) - (3n/2)

= (6n - 3n) / 2

3n/2 = 3n/2

LHS = RHS

THEREFORE S3 = 3(S2 - S1) !

HENCE PROVED !