Let T_r be the r^{text {th }} term of an A.P. | vedclass

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Question

$T_{m}=\frac{1}{25}, T_{25}=\frac{1}{20}, 20 \sum_{r=1}^{25} T_{r}=13$
$T_{m}=a+(m-1) d=\frac{1}{25} \ldots \ldots . .(1)$
$T_{25}=a+24 d=\frac{1}{2

Vedclass · Accepted Answer

$126$

Vedclass · Answer

$T_{m}=\frac{1}{25}, T_{25}=\frac{1}{20}, 20 \sum_{r=1}^{25} T_{r}=13$
$T_{m}=a+(m-1) d=\frac{1}{25} \ldots \ldots . .(1)$
$T_{25}=a+24 d=\frac{1}{20}$
$20 \cdot \frac{25}{2}\left[a+\frac{1}{20}ight]=13 \Rightarrow a=\frac{1}{500}$
$	ext { also, } 20 S_{25}=20 \cdot \frac{25}{2}[2 a+24 d]=13 \Rightarrow d=\frac{1}{500}$
from (1) $\frac{1}{500}+\frac{ m -1}{500}=\frac{1}{25} \Rightarrow m=20$
Now,
$5 m \sum_{r=m}^{2 m} T_{r}=100 \sum_{r=20}^{40} T_{r}=126$

Let $T_r$ be the $r^{\text {th }}$ 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 $5 m \sum_{ r = m }^{2 m} T _{ r }$ is equal to:

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