Question

If the sum of the first $m$ terms of an AP be $n$ and the sum of its first $n$ terms be $m$ then show that the sum of its first term is $-(m+n)$.

Answer 1

Sum of m terms $\Rightarrow \frac{m}{2}(2a+(m-1\left)d\right)=n$ ……….$\left(1\right)$

Sum of n terms $\Rightarrow \frac{n}{2}(2a+(n-1\left)d\right)=m$ ………..$\left(2\right)$

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

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

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$d(m-1-n+1)=\frac{2n}{m}-\frac{2m}{n}$

$d(m-n)=2\left(\frac{n^2-m^2}{nm}\right)$

$d=\frac{-2(m+n)}{mn}$

From $\left(2\right)$

$\text{/}2a+(n-1)(-\frac{\text{/}2(m+n)}{mn})=\frac{\text{/}2m}{n}$

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

Sum of $(m+n)$ terms $\Rightarrow \frac{m+n}{2}[2\left[\frac{m^2+n^2-m-n+mn}{mn}\right]+[m+n-1\left]\left(\frac{-2(m+n)}{mn}\right)\right]$

$\Rightarrow \frac{m+n}{2}\left[\frac{2m^2+2n^2-2m-2n+2mn-2m^2-4mn-2n^2-2m-2n}{mn}\right]$

$\Rightarrow \frac{m+n}{2mn}(-2mn)$

$\Rightarrow -(m+n)$.