Let T_r be the r^ term of an A.P.

Question

Let ${T}_{r}$ be the ${r}^{}$ term of an A.P. If for some $m,{T}_{m}=\frac{1}{25},{T}_{25}=\frac{1}{20}$, and $20\sum _{r=1}^{25}{T}_{r}=13$, then $5m\sum _{r=m}^{2m}{T}_{r}$ is equal to

Accepted Answer

To solve this problem, we start by analyzing the terms of an arithmetic progression (A.P.) where:

${T}_{m}=\frac{1}{25}$

${T}_{25}=\frac{1}{20}$

$20\sum _{r=1}^{25}{T}_{r}=13$

The formula for the ${r}^{}$ term of an A.P. is:

${T}_{r}=a+(r-1)d$

Given:

${T}_{m}=a+(m-1)d=\frac{1}{25}\,$

${T}_{25}=a+24d=\frac{1}{20}$

The sum of the first 25 terms (${S}_{25}$) is given by:

${S}_{25}=\frac{25}{2}	imes (2a+24d)$

Substituting into the equation for the sum:

$20	imes \frac{25}{2}	imes (2a+24d)=13$

This simplifies to:

$a=\frac{1}{500}$

Substituting $a=\frac{1}{500}$ into ${T}_{25}=a+24d=\frac{1}{20}$, we find:

$\frac{1}{500}+24d=\frac{1}{20}$

$24d=\frac{1}{20}-\frac{1}{500}$

$d=\frac{1}{500}$

Using Equation 1 again:

$\frac{1}{500}+\frac{m-1}{500}=\frac{1}{25}$

$\frac{m}{500}=\frac{1}{25}$

$m=20$

Now to find $5m\sum _{r=m}^{2m}{T}_{r}$:

$5m=5	imes 20=100$

$\sum _{r=20}^{40}{T}_{r}=\frac{21}{2}	imes ({T}_{20}+{T}_{40})$

Since $m=20$, the summation covers terms from ${T}_{20}$ to ${T}_{40}$, and:

$100	imes \sum _{r=20}^{40}{T}_{r}=126$

Therefore, $5m\sum _{r=m}^{2m}{T}_{r}=126$.

Solutions · 1