Let Tr be the rth term of an A.P. whose first term is a and common difference is d. If for some posi

Q: Wait, can I just pick $m=2$ and $n=3$ to solve this? It seems much faster than doing the algebra.

You have the heart of a true JEE strategist! Yes, in a high-pressure exam environment, substituting values like $m=2$ and $n=3$ is a brilliant 'trap-catcher' technique. If $T_2 = 1/3$ and $T_3 = 1/2$, you can quickly find $d$ and $a$ and verify the answer. However, as a student of physics and mathematics, we must also master the general proof. The algebraic method ensures that our intuition holds for any $m$ and $n$, not just the ones we picked.

Q: Why does the $a$ term disappear so cleanly when we subtract the equations?

That is the beauty of linearity! Because an Arithmetic Progression is essentially a linear function of the index $r$, the 'starting value' $a$ acts like a constant offset. When you subtract two terms of the same progression, that constant offset cancels out, leaving only the difference caused by the common difference $d$ and the change in index. It is the same principle as finding the slope of a line $y = mx + c$ by calculating $\frac{y_2 - y_1}{x_2 - x_1}$—the intercept $c$ always vanishes.

Q: Is it always true that $a=d$ when $T_m=1/n$ and $T_n=1/m$?

In this specific configuration, yes! It is a beautiful property of this reciprocal relationship. When the $m^{th}$ term is the reciprocal of $n$ and the $n^{th}$ term is the reciprocal of $m$, the progression is forced to pass through the origin $(0,0)$ if we were to extrapolate backwards. This implies the y-intercept $(a-d)$ must be zero, which mathematically forces $a = d$.

Question

Let $T_r$ be the $r$th term of an A.P. whose first term is $a$ and common difference is $d$. If for some positive integers $m, n, m 
eq n, T_m = 1/n$ and $T_n = 1/m$, then $a - d$ equals

Options:

(A) $\frac{1}{m} + \frac{1}{n}$

(B) $1$

(C) $\frac{1}{mn}$

(D) $0$

Sigma Percentile · Accepted Answer

### The Hidden Geometry of Arithmetic Progressions

Imagine you are standing on a coordinate plane. You have been given a sequence of numbers, an Arithmetic Progression (AP), but I want you to stop thinking of it as a list.

Instead, visualize it as a set of points on a graph. If you plot the term index $r$ on the horizontal axis and the value of the term $T_r$ on the vertical axis, you are not just looking at a sequence; you are looking at a perfectly straight line.

This is the first secret of the AP: it is a linear function in disguise. The equation of this line is:

$$T_r = dr + (a-d)$$

Here, $d$ is the slope, and $(a-d)$ is the y-intercept. When we ask for $a-d$, we are essentially asking for the y-intercept of this line. Let us embark on this journey to find it.

### The Two Points of Truth

The problem provides us with two anchors: the $m^{th}$ term is $\frac{1}{n}$, and the $n^{th}$ term is $\frac{1}{m}$. In our coordinate system, these are two distinct points: $(m, \frac{1}{n})$ and $(n, \frac{1}{m})$.

We know the general formula for any term in an AP is $T_r = a + (r-1)d$. Let us translate our points into the language of algebra:

$$a + (m-1)d = \frac{1}{n}$$

$$a + (n-1)d = \frac{1}{m}$$

We now have a system of two linear equations. Do not be intimidated by the variables $m$ and $n$; treat them as constants and focus on the structure.

### The Elegant Cancellation

Now, we face the challenge: we need to find $a-d$. We have two equations and two unknowns, $a$ and $d$. The most efficient path forward is to eliminate $a$.

If we subtract the second equation from the first, the $a$ terms will vanish, leaving us with a beautiful simplification:

$$[a + (m-1)d] - [a + (n-1)d] = \frac{1}{n} - \frac{1}{m}$$

On the left side, the $a$ cancels out, and we are left with $(m-1-n+1)d$, which simplifies to $(m-n)d$. On the right side, we find a common denominator:

$$(m-n)d = \frac{m-n}{mn}$$

### The Final Revelation

We are left with the equation $(m-n)d = \frac{m-n}{mn}$. Since the problem guarantees that $m 
eq n$, we know that $(m-n)$ is not zero.

We can safely divide both sides by $(m-n)$, revealing that the common difference $d$ is exactly:

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

Now, we substitute this back into our first equation to find $a$. With $d = \frac{1}{mn}$, the equation $a + (m-1)d = \frac{1}{n}$ becomes:

$$a + (m-1)\left(\frac{1}{mn}ight) = \frac{1}{n}$$

Expanding this, we get $a + \frac{m}{mn} - \frac{1}{mn} = \frac{1}{n}$. Since $\frac{m}{mn}$ is just $\frac{1}{n}$, the terms cancel out perfectly, leaving us with $a = \frac{1}{mn}$.

Finally, we calculate $a-d$. Since both $a$ and $d$ are equal to $\frac{1}{mn}$, their difference is zero. We have arrived at our destination: the y-intercept of our line is zero, meaning the progression passes through the origin.

The answer is **0**.

Animated Solution for Mathematics - Sequence and Series: Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

Select Answer:

Visualized Solution

Visualizing the A.P.

The Given Data Points

Setting up the Equations

Eliminating $a$

Simplifying the Left Hand Side

Simplifying the Right Hand Side

Solving for $d$

Finding the First Term $a$

Expanding the Term

Simplifying to find $a$

Calculating $a - d$

The Sigma Insight: Arithmetic Progression (A.P.)

The Hidden Geometry of Arithmetic Progressions

The Two Points of Truth

The Elegant Cancellation

The Final Revelation

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