# If the sum of first n terms of an AP is half the sum of next n terms, then $\large\frac{S_{4n}}{S_{n}}$ equals

$(a)\;10\qquad(b)\;8\qquad(c)\;6\qquad(d)\;4$

Explanation : Given ,
$S_{2n}-S_{n}=2S_{n}$
$S_{2n}=3S_{n}$
$\large\frac{2n}{2}\;[2a+(n-1)d]=\large\frac{3n}{2}\;[2a+(n-1)d]$
$4a+4nd-2d=6a+3nd-3d$
$2a=nd+d=(n-1)d$
$\large\frac{S_{4n}}{S_{n}}=\large\frac{\large\frac{4n}{2}\;[2a+(4n-1)d]}{\large\frac{n}{2}\;[2a+(n-1)d]}$
$\large\frac{S_{4n}}{S_{n}}=\large\frac{4\;[(n+1)d+(4n-1)d]}{[(n+1)d+(n-1)d]}$
$\large\frac{S_{4n}}{S_{n}}=\large\frac{4(5n)}{(2n)}=10\;.$