Question

show that s³ = 3 ( s2 - s1 )​

Answer 1

Answer:

If s1,s2 and s3 are the sum of first n,2n,and 3n terms of an A.P. respectively then,

s1 = n/2[2a + (n-1)d]

s2 = 2n/2[2a + (2n-1)d]

s3 = 3n/2[2a + (3n-1)d]

consider that,

s2 - s1 = 2n/2[2a + (n-1)d] - n/2[2a + (n-1)d]

          = n/2 {[4a + 2(2n - 1)d] - [2a + (n-1)d]

s2 - s1 =n/2[2a +(3n - 1)d]

3(s2 - s1) = 3 x n/2[2a + (3n - 1)d]

3(s2 - s1) = 3n/2[2a + (3n - 1)d]

3(s2 - s1) = s3 { ∵ s3 = 3n/2[2a + (3n-1)d]}

hence, proved.

Answer 2

Answer:

Answer in attachment.

Hope it will help you .