Question

prove that the sum of the latter half of 2n terms of any ap is one third the the sum of 3n terms of the same AP

Answer 1

Sum of an AP sequence is : Sn = (n/2)[2a + (n-1)d]

To find the sum of latter half of 2n terms of AP we need to find : S2n - Sn

Sn = (n/2)[2a + (n-1)d] = an + n²d/2 - nd/2

S2n = (2n/2)[2a + (2n-1)d] = 2an + 2n²d - nd

S2n - Sn = 2an + 2n²d - nd - an - n²d/2 + nd/2

S2n - Sn = an + 3n²d/2 - nd/2

Lets find S3n :

S3n = 3n/2[2a + (3n - 1)d]

S3n = 3an + 9n²d/2 - 3nd/2

(1/3)S3n = an + 3n²d/2 - nd/2

which is equal to S2n - Sn

Therefor S2n - Sn = (1/3)S3n