Question

The common difference of an arithmetic sequence is d prove that the sum of first n terms and next n terms differ by n×n×d

Answer 1

let's test it with an example:
let a = 2 , d = 3 , n = 10
sum(10) = 5(4 + 27) = 155
sum(20) = 10(4 + 57) = 610
so first 10 terms = 155 , next 10 terms = 610-155 = 455
difference = 300
= 10x10x3 , looks like it works

Sum(n) = (n/2)(2a + (n-1)d)
sum(2n) = (2n/2)(2a + (2n-1)d) = n(2a + (2n-1)d)
so the next n terms = n(2a + (2n-1)d) - (n/2)(2a + (n-1)d)
= (n/2)(2a + 3dn - d) , leaving it up to you to have found that



the difference between the first n terms and the next n terms

= (n/2)(2a + 3dn - d) - (n/2)(2a + nd - d) , common factor of n/2
= (n/2)[ 2a + 3dn - d - 2a - nd + d]
= (n/2)(2nd)
= n^2 d
= n x n x d as required