Question

Prove that the sum of the latter half of $2n$ terms of an AP is equal to one-third of the sum of the first $3n$ terms.

Answer 1

To prove $S_{2n}-S_n=\frac{1}{3}{S}_{3n}$:

Sum of later half of $2n$ terms of AP $=S_{2n}-S_n$:

Formula used:

We know that the sum of the first $n$ terms of an AP is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Where $a=$ first term and $d=$ common difference.

$\begin{array}{rcl}{S}_{2n}-S_n& =& \frac{n}{2}[2a+(2n-1\left)d\right]-\frac{n}{2}[2a+(n-1\left)d\right]\\ & =& \frac{n}{2}[4a+2(2n-1)d-2a-(n-1\left)d\right]\\ & =& \frac{n}{2}[4a+4nd-2d-2a-nd+d]\\ & =& \frac{n}{2}[2a+(4n-2-n+1\left)d\right]\\ & =& \frac{n}{2}[2a+(3n-1\left)d\right]\\ & =& \frac{1}{3}\times \frac{3n}{2}[2a+(3n-1\left)d\right]\\ & =& \frac{1}{3}{S}_{3n}\\ & =& \frac{1}{3}(\mathrm{Sum}\mathrm{of}\mathrm{first}3n\mathrm{terms})\end{array}$

Therefore, the sum of later half of $2n$ terms of an AP is equal to one-third of the sum of the first $3n$ terms.

Hence proved.