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

Question

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 $20 \sum\limits_{\mathrm{r}=1}^{25} \mathrm{~T}_{\mathrm{r}}=13$, then $5 \mathrm{~m} \sum\limits_{\mathrm{r}=\mathrm{m}}^{2 \mathrm{~m}} \mathrm{~T}_{\mathrm{r}}$ is equal to

Accepted Answer

To solve this problem, we start by analyzing the terms of an arithmetic progression (A.P.) where:

$ T_m = \frac{1}{25} $

$ T_{25} = \frac{1}{20} $

$ 20 \sum\limits_{r=1}^{25} T_r = 13 $

The formula for the $ r^{	ext{th}} $ term of an A.P. is:

$ T_r = a + (r-1)d $

Given:

$ T_m = a + (m-1)d = \frac{1}{25} \quad 	ext{(Equation 1)} $

$ T_{25} = a + 24d = \frac{1}{20} $

The sum of the first 25 terms ($ S_{25} $) is given by:

$ S_{25} = \frac{25}{2} 	imes (2a + 24d) $

Substituting into the equation for the sum:

$ 20 	imes \frac{25}{2} 	imes (2a + 24d) = 13 $

This simplifies to:

$ a = \frac{1}{500} $

Substituting $ a = \frac{1}{500} $ into $ T_{25} = a + 24d = \frac{1}{20} $, we find:

$ \frac{1}{500} + 24d = \frac{1}{20} $

$ 24d = \frac{1}{20} - \frac{1}{500} $

$ d = \frac{1}{500} $

Using Equation 1 again:

$ \frac{1}{500} + \frac{m-1}{500} = \frac{1}{25} $

$ \frac{m}{500} = \frac{1}{25} $

$ m = 20 $

Now to find $ 5m \sum\limits_{r=m}^{2m} T_r $:

$ 5m = 5 	imes 20 = 100 $

$ \sum\limits_{r=20}^{40} T_r = \frac{21}{2} 	imes (T_{20} + T_{40}) $

Since $ m = 20 $, the summation covers terms from $ T_{20} $ to $ T_{40} $, and:

$ 100 	imes \sum\limits_{r=20}^{40} T_r = 126 $

Therefore, $ 5m \sum\limits_{r=m}^{2m} T_r = 126 $.

Solutions · 1