Question

The sum of the latter half of $2n$ terms of any A.P. is one-third the sum of $3n$ terms of the same A.P.

• A. True
• B. False

Answer 1

The correct option is A True

Sum of later half of the $2n$ terms means sum of last n terms which is equal to

$=\frac{2n}{2}(2a+(2n-1\left)d\right)-\frac{n}{2}(2a+(n-1)d$

On taking $\frac{n}{2}$ common

$=\frac{n}{2}(4a+2(2n-1)d-2a-(n-1\left)d\right))$

$=\frac{n}{2}(2a+2(2n-1)d-(n-1\left)d\right))$

$=\frac{n}{2}(2a+d(4n-2-n+1)$

$=\frac{n}{2}(2a+(3n-1\left)d\right)=$

$=\frac{1}{3}\left[\frac{3n}{2}\right(2a+(3n-1)d\left)\right]$

So, one-third the sum of first $3n$ terms

Hence, this is the answer.