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

<p>To solve this problem, we start by analyzing the terms of an arithmetic progression (A.P.) where:</p> <p><p>$ T_m = \frac{1}{25} $</p></p> <p><p>$ T_{25} = \frac{1}{20} $</p></p> <p><p>$ 20 \sum\limits_{r=1}^{25} T_r = 13 $</p></p> <p>The formula for the $ r^{\text{th}} $ term of an A.P. is:</p> <p>$ T_r = a + (r-1)d $</p> <p>Given:</p> <p>$ T_m = a + (m-1)d = \frac{1}{25} \quad \text{(Equation 1)} $</p> <p>$ T_{25} = a + 24d = \frac{1}{20} $</p> <p>The sum of the first 25 terms ($ S_{25} $) is given by:</p> <p>$ S_{25} = \frac{25}{2} \times (2a + 24d) $</p> <p>Substituting into the equation for the sum:</p> <p>$ 20 \times \frac{25}{2} \times (2a + 24d) = 13 $</p> <p>This simplifies to:</p> <p>$ a = \frac{1}{500} $</p> <p>Substituting $ a = \frac{1}{500} $ into $ T_{25} = a + 24d = \frac{1}{20} $, we find:</p> <p>$ \frac{1}{500} + 24d = \frac{1}{20} $</p> <p>$ 24d = \frac{1}{20} - \frac{1}{500} $</p> <p>$ d = \frac{1}{500} $</p> <p>Using Equation 1 again:</p> <p>$ \frac{1}{500} + \frac{m-1}{500} = \frac{1}{25} $</p> <p>$ \frac{m}{500} = \frac{1}{25} $</p> <p>$ m = 20 $</p> <p>Now to find $ 5m \sum\limits_{r=m}^{2m} T_r $:</p> <p>$ 5m = 5 \times 20 = 100 $</p> <p>$ \sum\limits_{r=20}^{40} T_r = \frac{21}{2} \times (T_{20} + T_{40}) $</p> <p>Since $ m = 20 $, the summation covers terms from $ T_{20} $ to $ T_{40} $, and:</p> <p>$ 100 \times \sum\limits_{r=20}^{40} T_r = 126 $</p> <p>Therefore, $ 5m \sum\limits_{r=m}^{2m} T_r = 126 $.</p>