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

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(B) Given $T_m = a + (m-1)d = \frac{1}{25}$ and $T_{25} = a + 24d = \frac{1}{20}$.Sum of $A.P.$ is $S_n = \frac{n}{2}(a + T_n)$.Given $20 \sum_{r=1}^{

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$126$

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(B) Given $T_m = a + (m-1)d = \frac{1}{25}$ and $T_{25} = a + 24d = \frac{1}{20}$.Sum of $A.P.$ is $S_n = \frac{n}{2}(a + T_n)$.Given $20 \sum_{r=1}^{25} T_r = 13 \Rightarrow 20 	imes \frac{25}{2}(a + T_{25}) = 13$.$250(a + \frac{1}{20}) = 13$ $\Rightarrow a + \frac{1}{20} = \frac{13}{250}$ $\Rightarrow a = \frac{13}{250} - \frac{1}{20} = \frac{26-25}{500} = \frac{1}{500}$.Substitute $a$ in $T_{25} = a + 24d = \frac{1}{20}$ $\Rightarrow \frac{1}{500} + 24d = \frac{25}{500}$ $\Rightarrow 24d = \frac{24}{500}$ $\Rightarrow d = \frac{1}{500}$.From $T_m = a + (m-1)d = \frac{1}{25}$ $\Rightarrow \frac{1}{500} + \frac{m-1}{500} = \frac{20}{500}$ $\Rightarrow m-1 = 19$ $\Rightarrow m = 20$.We need to find $5m \sum_{r=m}^{2m} T_r = 100 \sum_{r=20}^{40} T_r$.Sum $= \frac{n}{2}(T_{first} + T_{last}) = \frac{40-20+1}{2}(T_{20} + T_{40}) = \frac{21}{2}(a+19d + a+39d) = \frac{21}{2}(2a+58d) = 21(a+29d)$.$21(\frac{1}{500} + \frac{29}{500}) = 21(\frac{30}{500}) = 21(\frac{3}{50}) = \frac{63}{50} = 1.26$.$100 	imes 1.26 = 126$.

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 $5m \sum_{r=m}^{2m} T_r$ is equal to:

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