Question

If the sum of $m$ terms of an AP is equal to the sum of either the next $n$ terms or the next $p$ terms, then prove that $(m+n)(\frac{1}{m}-\frac{1}{p})=(m+p)(\frac{1}{m}-\frac{1}{n})$.

Answer 1

Let's start arbitrarily with the term $a\left(1\right)$.

Sum of first $m$ terms

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

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

Sum of next $n$ terms

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

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

Sum of next $p$ terms

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

$=\frac{p}{2}\left[2a\right(1)+(2m+2n+p-1\left)d\right]$ $.............\left(3\right)$

All of these equal $S$. So, now we need to eliminate $S$ and $a\left(1\right)$ and $d$.

$S(\frac{2}{n}-\frac{2}{m})=(m+n)d$ $............\left(4\right)$

$S(\frac{2}{p}-\frac{2}{m})=(m+p)d$ $.............\left(5\right)$

On dividing, we get

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

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

$(m+n)(\frac{1}{m}-\frac{1}{p})-(m+p)(\frac{1}{m}-\frac{1}{n})=0$

Hence, this is the answer.