Let $\mathrm{T}_{\mathrm{r}}$ be the $\mathrm{r}^{ ext {th }}$ term of an A.P. If for some $\mathrm{m}, \mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}_{25}=\frac{1}{20}$, and $20 \sum_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13$, then $5 \mathrm{~m} \sum_{\mathrm{r}=\mathrm{m}}^{2…

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

Aicharya · Accepted Answer

$\begin{aligned}
& \mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}_{25}=\frac{1}{20}, 20 \sum_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13 \\ & \mathrm{~T}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}=\frac{1}{25} \ldots \ldots .(1) \\ & \mathrm{T}_{25}=\mathrm{a}+24 \mathrm{~d}=\frac{1}{20} \\ & 20 \cdot \frac{25}{2}\left[\mathrm{a}+\frac{1}{20}\right]=13 \Rightarrow \mathrm{a}=\frac{1}{500}
\end{aligned}$
also, $20 \mathrm{~S}_{25}=20 \cdot \frac{25}{2}[2 \mathrm{a}+24 \mathrm{~d}]=13 \Rightarrow \mathrm{~d}=\frac{1}{500}$
from (1) $\frac{1}{500}+\frac{\mathrm{m}-1}{500}=\frac{1}{25} \Rightarrow \mathrm{~m}=20$
Now,
$5 \mathrm{~m} \sum_{\mathrm{r}=\mathrm{m}}^{2 \mathrm{~m}} \mathrm{~T}_{\mathrm{r}}=100 \sum_{\mathrm{r}=20}^{40} \mathrm{~T}_{\mathrm{r}}=126$ *

Let $\mathrm{T}_{\mathrm{r}}$ be the $\mathrm{r}^{\text {th }}$ term of an A.P. If for some $\mathrm{m},…

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