# For an AP with first term a and sum of first m terms 0, the sum of next n terms will be

$(a)\;\frac{a(m-n)n}{m-1}\qquad(b)\;\frac{-a(m+n)n}{m-1}\qquad(c)\;\frac{a(m+n)n}{m-1}\qquad(d)\;None\;of\;these$

Answer : (a) $\;\frac{a(m-n)n}{m-1}$
Explanation : Given
$\frac{m}{2}\;[2a+(m-1)d]=0$
$d=\frac{-2a}{m-1}$
$=\frac{n}{2}\;[2a+(n-1)d]$
$=\frac{n}{2}\;[2a-\frac{2a(n-1)}{m-1}]$
$=\frac{a(m-n)n}{m-1}\;.$
answered Jan 21, 2014 by