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The sum of first $n$ terms of an $A.P$. whose first term is $\pi$ is zero. The sum of next $m$ terms is :

- $\frac{ \pi m \left(m+n \right)}{n-1}$
- $\frac{ \pi n \left(m+n \right)}{1-n}$
- $\frac{ \pi m \left(m+n \right)}{1-n}$
- $1$

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Some formulas which am going to use in answering this question:

If $’a’$ is the first term and $’d’$ is the common difference then,

- $n^{th}$ of an A.P. sequence is given by $T_n = a+(n-1)d$
- Sum of $n$ terms of an A.P. sequence is given by $S_n = \cfrac{n}{2}\Bigl[2a+(n-1)d\Bigr]$

Given,

The sum of first $n$ terms of an A.P. whose first term is $\pi$ is zero.

$(\pi) + (\pi+d)+ (\pi+2d)+ (\pi+3d) +\cdots+\Bigl(\pi+(n-1)d\Bigr) = 0$

$\cfrac{n}{2}\Bigl(2(\pi)+(n-1)d\Bigr)=0$$\require{cancel}$

$2n{\pi} +dn(n-1)=0$

$d=\cfrac{-2\cancel{n}\pi}{\cancel{n}(n-1)}$

$d=\cfrac{-2\pi}{(n-1)}$

Now the question is asking

The sum of next m terms is :

Here first term $’a’$ becomes $(\pi+nd)$ for the next $’m’$ terms.

Let $S_m$ denote the sum of next $’m’$ terms.

$S_m = (\pi+nd) + (\pi+nd+d)+ (\pi+nd+2d)+ (\pi+nd+3d) +\cdots+\Bigl(\pi+nd+(m-1)d\Bigr)$

$S_m = \cfrac{m}{2}\Bigl[2(\pi +nd)+(m-1)d\Bigr]$

$S_m = m(\pi+nd)+\cfrac{md(m-1)}{2}$

$S_m = m\pi+mnd+\cfrac{m^2d-md}{2}$

$S_m = m\pi+\cfrac{d}{2}(m^2-m+2mn)$

$S_m = m\pi+\cfrac{\left(\cfrac{\cancel{-2} \pi}{n-1}\right) }{\cancel{2}}(m^2-m+2mn)$

$S_m = m\pi+\cfrac{-\pi}{n-1}(m^2-m+2mn)$

$S_m = \cfrac{m\pi(n-1)-m^2\pi+m\pi-2mn\pi}{n-1}$

$S_m = \cfrac{m\pi n\cancel{-m\pi}-m^2\pi\cancel{+m\pi}-2mn\pi}{n-1}$

$S_m = \cfrac{-m\pi n-m^2\pi}{n-1}$

$S_m = \cfrac{-m\pi (n+m)}{-(1-n)}$

$S_m = \cfrac{\pi m(m+n)}{(1-n)}$

$\therefore$ Option C is the correct answer.