Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by $d$.
• r-th term of an AP: $T_r = a + (r-1)d$, where $a$ is the first term.
• Sum of the first n terms of an AP: $S_n = \sum_{r=1}^{n} T_r = \frac{n}{2}[2a + (n-1)d]$.
• Sum of a block of terms in an AP: The sum of terms from $T_m$ to $T_{2m}$ is the sum of an AP with $(2m - m + 1) = m+1$ terms, where the first term is $T_m$ and the last term is $T_{2m}$. The sum is $\frac{m+1}{2}[T_m + T_{2m}]$.

Step-by-Step Solution

We are given an Arithmetic Progression (AP) with terms $T_r$. We have the following information:

1. $T_m = \frac{1}{25}$
2. $T_{25} = \frac{1}{20}$
3. $20 \sum_{r=1}^{25} T_r = 13$

Our goal is to find the value of $5m \sum_{r=m}^{2m} T_r$.

Step 1: Expressing the given information using AP formulas. Let $a$ be the first term and $d$ be the common difference of the AP. From the definition of the $r^{th}$ term: $T_m = a + (m-1)d = \frac{1}{25}$ (Equation 1) $T_{25} = a + (25-1)d = a + 24d = \frac{1}{20}$ (Equation 2)

The sum of the first 25 terms is $\sum_{r=1}^{25} T_r = S_{25}$. Using the sum formula: $S_{25} = \frac{25}{2}[2a + (25-1)d] = \frac{25}{2}(2a + 24d)$

Now, substitute this into the third given condition: $20 \times \frac{25}{2}(2a + 24d) = 13$ $250(2a + 24d) = 13$ $2a + 24d = \frac{13}{250}$ (Equation 3)

Step 2: Solving for the first term ($a$) and the common difference ($d$). We have a system of equations. Let's use Equation 2 and Equation 3. From Equation 2: $a + 24d = \frac{1}{20}$ From Equation 3: $2a + 24d = \frac{13}{250}$

Subtract Equation 2 from Equation 3: $(2a + 24d) - (a + 24d) = \frac{13}{250} - \frac{1}{20}$ $a = \frac{13}{250} - \frac{12.5}{250}$ (Converting $\frac{1}{20}$ to $\frac{12.5}{250}$) $a = \frac{0.5}{250} = \frac{1}{500}$

Now substitute the value of $a$ into Equation 2 to find $d$: $\frac{1}{500} + 24d = \frac{1}{20}$ $24d = \frac{1}{20} - \frac{1}{500}$ $24d = \frac{25}{500} - \frac{1}{500}$ $24d = \frac{24}{500}$ $d = \frac{1}{500}$

So, the first term $a = \frac{1}{500}$ and the common difference $d = \frac{1}{500}$.

Step 3: Finding the value of m. We use Equation 1: $a + (m-1)d = \frac{1}{25}$ Substitute the values of $a$ and $d$: $\frac{1}{500} + (m-1)\frac{1}{500} = \frac{1}{25}$ Multiply the entire equation by 500: $1 + (m-1) = \frac{500}{25}$ $1 + m - 1 = 20$ $m = 20$

Step 4: Calculating the required expression $5m \sum_{r=m}^{2m} T_r$. We need to calculate $5m \sum_{r=m}^{2m} T_r$. Since $m=20$, we need to calculate $5 \times 20 \sum_{r=20}^{2 \times 20} T_r = 100 \sum_{r=20}^{40} T_r$.

The summation $\sum_{r=20}^{40} T_r$ represents the sum of terms from $T_{20}$ to $T_{40}$. This is an AP with $40 - 20 + 1 = 21$ terms. The first term of this summation is $T_{20}$ and the last term is $T_{40}$.

Let's calculate $T_{20}$ and $T_{40}$ using $a = \frac{1}{500}$ and $d = \frac{1}{500}$: $T_{20} = a + (20-1)d = \frac{1}{500} + 19 \times \frac{1}{500} = \frac{1+19}{500} = \frac{20}{500} = \frac{1}{25}$ $T_{40} = a + (40-1)d = \frac{1}{500} + 39 \times \frac{1}{500} = \frac{1+39}{500} = \frac{40}{500} = \frac{2}{25}$

Now, calculate the sum $\sum_{r=20}^{40} T_r$: $\sum_{r=20}^{40} T_r = \frac{\text{Number of terms}}{2} \times (\text{First term} + \text{Last term})$ $\sum_{r=20}^{40} T_r = \frac{21}{2} (T_{20} + T_{40})$ $\sum_{r=20}^{40} T_r = \frac{21}{2} \left(\frac{1}{25} + \frac{2}{25}\right)$ $\sum_{r=20}^{40} T_r = \frac{21}{2} \left(\frac{3}{25}\right) = \frac{63}{50}$

Finally, calculate the required expression: $5m \sum_{r=m}^{2m} T_r = 100 \times \sum_{r=20}^{40} T_r$ $= 100 \times \frac{63}{50}$ $= 2 \times 63 = 126$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with fractions and signs. A small error can lead to a completely wrong answer.
• Summation Range: Ensure the number of terms in a summation is calculated correctly. For a sum from $T_m$ to $T_{2m}$, there are $(2m - m + 1)$ terms.
• Using the Correct Formulas: Always use the standard formulas for the $r^{th}$ term and the sum of an AP.

Summary The problem involves an arithmetic progression where we are given specific terms and a sum of terms. By expressing these conditions using the general formulas for an AP, we formed a system of equations to solve for the first term ($a$) and the common difference ($d$). Once $a$ and $d$ were found, we determined the value of $m$ using the given information about $T_m$. Finally, we calculated the required summation by finding the first and last terms of the specified range and applying the sum formula for an AP.

The final answer is $\boxed{126}$.