Question

The sum of the first n terms of an A. P. is 2n and the sum of the first 2n terms is n. Find the sum of the first 4n terms.

Answer 1

Answer:

The sum of first 4n terms is -10n.

Step-by-step explanation:

The formula for sum of n terms of an AP is

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

It is given that the sum of the first n terms of an A. P. is 2n and the sum of the first 2n terms is n.

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

$2a+(n-1)d=4$                          ..... (1)

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

$2a+(2n-1)d=1$                          ..... (2)

Solve equation (1) and (2) by elimination method.

$d=-\frac{3}{n}$

$a=\frac{1}{2}(7-\frac{3}{n})$

The sum of first 4n terms is

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

$S_{4n}=2n[2a+(4n-1)d]$

Put the value of a and d.

$S_{4n}=2n[2(\frac{1}{2}(7-\frac{3}{n}))+(4n-1)(-\frac{3}{n})]$

$S_{4n}=2n[7-\frac{3}{n}-12+\frac{3}{n}]$

$S_{4n}=2n[-5]$

$S_{4n}=-10n$

Therefore the sum of 4n terms is -10n.