Let T_r be the r-th term of an A.P.

Question

Let $T_r$ be the $r$-th term of an A.P. If $T_m = \dfrac{1}{25}$ for some $m$, $T_{25} = \dfrac{1}{20}$, and $\displaystyle\sum_{r=1}^{25}T_r = 13$, then $5m\displaystyle\sum_{r=m}^{2m}T_r$ is equal to:

Accepted Answer

$S_{25} = \dfrac{25}{2}(2a+24d) = 13 \Rightarrow a+12d = \dfrac{13}{25}$.

$T_{25} = a+24d = \dfrac{1}{20}$. Solving these two equations gives $a=\dfrac{1}{500}$, $d=\dfrac{1}{500}$.

From $T_m = a+(m-1)d = \dfrac{1}{25}$: $\dfrac{1}{500}+\dfrac{m-1}{500}=\dfrac{1}{25} \Rightarrow m = 20$.

$$\sum_{r=20}^{40}T_r = \frac{21}{2}(T_{20}+T_{40}) = 1.26, \qquad 5m\sum_{r=m}^{2m}T_r = 100\times1.26 = 126.$$

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