Question

Prove that the sum of later half of 2n terms of an ap is equal to one-third of the sum of the first 3n terms.

Answer 1

sum of Later half  of the 2n terms means sum of last n terms which is equal to 
$S_{2n}$-$S_{n}$
2n/2(2a+2n-1)d)-n/2(2a+(n-1)d
taking n/2 common 
n/2(2a+2(2n-2)d-2a-(n-1)d))
=n/2(2a+d(4n-2-n+1)
= n/2(2a+(3n-1)d)= Sum of first 3n terms 

Answer 2

REFER TO THE ATTACHMENT.......

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