Question

The first term of an $A.P.isa$, and the sum of its first $n$ term is zero. Then sum of next $m$ terms=

• A. $an.\frac{m-n}{1-m}$
• B. $am.\frac{m-n}{1-n}$
• C. $an.\frac{m+n}{1-m}$
• D. $am.\frac{m+n}{1-n}$

Answer 1

The correct option is D $am.\frac{m+n}{1-n}$

first term $=a$

sum of first $n$ term $0$

$\frac{n}{2}[2a+(n-1\left)d\right]=0$

$d=\frac{-2a}{n-1}$

$(n+1)$ term $=a+nd$

and $nd$ will be $1^{st}$ term for next $m$ terms

$s_m=\frac{m}{2}\left[2\right(a+nd)+(m-1\left)d\right]$

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

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

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

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

$=am\left[\frac{n-1-2n-m+1}{n-1}\right]$

$S_m=am\left[\frac{m+n}{1-n}\right]$