Animated Solution for Mathematics - Sequence and Series: Let $T_r$ be the rth term of an A.P. If for some m, $T_m=\frac{1}{25}$ and $T_{25}=\frac{1}{20}$ , then $5m{\sum }_{r=m}^{2m}{T}_r$ is equal to Note: $20{\sum }_{r=1}^{25}{T}_r=13$ is given condition in Hindi version.

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of an Arithmetic Progression (AP) to see the elegant machinery underneath.

When you look at a problem involving $T_r$ and $\sum T_r$, do not see a wall of text. See the DNA of the sequence. Every AP is governed by two fundamental constants: the first term $a$ and the common difference $d$.

If you know these two, you know everything about the sequence. Our mission is to hunt them down.

Decoding the DNA

We are given $T_m=\frac{1}{25}$ and $T_{25}=\frac{1}{20}$. Immediately, we translate these into the language of algebra.

The general term is $T_n=a+(n-1)d$. Thus, for the 25th term, we have the equation:

This is our anchor. Now, consider the summation condition: $20{\sum }_{r=1}^{25}{T}_r=13$.

The sum of the first $n$ terms is given by $S_n=\frac{n}{2}[2a+(n-1)d]$. For $n=25$, this becomes:

The Elegant Substitution

Here is where the magic happens. Many students would blindly expand $2a+24d$. But look closer; we have $2a+24d$, and we know that $a+24d=\frac{1}{20}$.

Let us rewrite the sum expression:

Substituting our known value, we get $S_{25}=\frac{25}{2}[a+\frac{1}{20}]$. Now, applying the condition $20S_{25}=13$:

With $a$ in our pocket, finding $d$ is trivial. Plugging $a$ back into $a+24d=\frac{1}{20}$, we find $d=\frac{1}{500}$. We have successfully decoded the DNA of this sequence.

The Hunt for $m$

Now that we have $a$ and $d$, finding $m$ is just a matter of walking the path. We know $T_m=\frac{1}{25}$.

Expanding this:

Solving this gives us $m=20$. We are halfway to victory.

The Final Summation

We are asked to find $5m{\sum }_{r=m}^{2m}{T}_r$. With $m=20$, this becomes $100{\sum }_{r=20}^{40}{T}_r$.

Instead of calculating each term, we use the property that the sum of an AP is the number of terms multiplied by the average of the first and last terms. The number of terms from $r=20$ to $r=40$ is $40-20+1=21$.

Using our values $T_{20}=\frac{20}{500}$ and $T_{40}=\frac{40}{500}$, the sum is:

Finally, multiplying by 100:

And there it is. 126. The beauty of this problem is not in the arithmetic, but in the structure. By identifying the parameters early and using the symmetry of the sum formula, we turned a complex-looking expression into a simple, elegant result.