Animated Solution for Mathematics - Sequence and Series: Let $T_r$ be the $r$th term of an A.P., for $r=1,2,3,\dots$. If for some positive integers $m,n$ we have $T_m=1/n$ and $T_n=1/m$, then $T_{mn}$ equals

Imagine you are standing at the edge of a vast, orderly landscape. This is the world of Arithmetic Progressions (AP), where everything moves with a constant, rhythmic step.

We are given two clues: the $m$-th term is $\frac{1}{n}$, and the $n$-th term is $\frac{1}{m}$. The indices $m$ and $n$ are swapped, and the values $\frac{1}{n}$ and $\frac{1}{m}$ are also swapped, revealing a beautiful mirror reflection.

To navigate this landscape, we use the general term formula for an AP: Here, $a$ is the first term and $d$ is the common difference. This formula serves as the bridge between abstract indices and concrete values.

We translate our clues into the language of algebra: 1. $a+(m-1)d=\frac{1}{n}$ 2. $a+(n-1)d=\frac{1}{m}$

We now have a system of two linear equations with two unknowns.

To isolate our variables, we subtract the second equation from the first. This causes the $a$ terms to vanish:

Simplifying the left side, the $-1$ and $+1$ cancel out, while the right side is combined using a common denominator:

Since $m$ and $n$ are distinct positive integers, $m-n eq 0$. We can safely divide both sides by $(m-n)$ to find the common difference:

Now, we substitute $d=\frac{1}{mn}$ into our first equation to find $a$:

Expanding this, we get:

Since $\frac{m}{mn}=\frac{1}{n}$, the terms cancel out, leaving us with:

We have discovered that $a=d=\frac{1}{mn}$. Our sequence is simply the sequence of multiples of $\frac{1}{mn}$.

To find the $mn$-th term, $T_{mn}$, we use the general formula:

Substituting our values:

Expanding the expression:

The $\frac{1}{mn}$ and $-\frac{1}{mn}$ terms cancel out perfectly, leaving us with the final result: $T_{mn}=1$

This is a standard gem in the JEE treasure chest. You have successfully mastered the symmetry of the arithmetic progression.