Question

If the sum of $m$ terms of an arithmetical progression is equal to the sum of either the next $n$ terms or the next $p$ terms, then $\left(\frac{n+m}{n-m}\right)\left(\frac{p-m}{p+m}\right)$ is-

• A. $\frac{n}{p}$
• B. $\frac{p}{n}$
• C. $np$
• D. $\frac{p}{m}$

Answer 1

Answer: (B)

$\frac{p}{n}$

Let the A.P. is given by $a_1,a_2,a_3,...$

Given $a_1+a_2+a_3+....+a_m=a_{m+1}+a_{m+2}+.....+a_{m+n}$

$\Rightarrow a_1+a_2+a_3....+a_m+a_1+a_2+a_3+...+a_m=a_1+a_2+....+a_m+a_{m+1}+a_{m+2}+...+a_{m+n}$

$\Rightarrow 2S_m=S_{m+n}$

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

$\Rightarrow (m-n)[2a+(m-1\left)d\right]=(m+n)nd$ .....(1)

Also given

$a_1+a_2+a_3+....+a_m=a_{m+1}+a_{m+2}+.....+a_{m+p}$

$\Rightarrow a_1+a_2+a_3....+a_m+a_1+a_2+a_3+...+a_m=a_1+a_2+....+a_m+a_{m+1}+a_{m+2}+...+a_{m+p}$

$\Rightarrow 2S_m=S_{m+p}$

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

$\Rightarrow (m-p)[2a+(m-1\left)d\right]=(m+p)pd$ .....(2)

Dividing (1) by (2), we get

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

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

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