Understanding the Arithmetic Progression Problem

This question involves an Arithmetic Progression (AP). An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, usually denoted by \(d\).

The \(r\)-th term of an AP is given by the formula:

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

where \(a\) is the first term and \(d\) is the common difference.

Setting up the Equations from Given Information

We are given that for two distinct positive integers \(m\) and \(n\):

• The \(m\)-th term \(T_m\) is equal to \(1/n\).
• The \(n\)-th term \(T_n\) is equal to \(1/m\).

Using the formula for the \(r\)-th term, we can write these two conditions as equations:

Equation 1: \(T_m = a + (m-1)d = \frac{1}{n}\)

Equation 2: \(T_n = a + (n-1)d = \frac{1}{m}\)

Our goal is to find the value of the \(mn\)-th term, \(T_{mn}\), which is \(T_{mn} = a + (mn-1)d\).

Finding the Common Difference (\(d\))

To find \(d\), we can subtract Equation 2 from Equation 1. This will eliminate the first term \(a\).

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

Expanding the terms:

\(a + md - d - a - (nd - d) = \frac{m - n}{mn}\)

\(a + md - d - a - nd + d = \frac{m - n}{mn}\)

The \(a\) and \(-d\) terms cancel out:

\(md - nd = \frac{m - n}{mn}\)

Factor out \(d\) from the terms on the left side:

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

Since \(m\) and \(n\) are distinct, \(m-n \neq 0\). We can divide both sides by \((m-n)\):

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

\(d = \frac{1}{mn}\)

Finding the First Term (\(a\))

Now that we have the value of \(d\), we can substitute it into either Equation 1 or Equation 2 to find \(a\). Let's use Equation 1:

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

Substitute \(d = \frac{1}{mn}\):

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

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

Subtract \(\frac{m-1}{mn}\) from both sides:

\(a = \frac{1}{n} - \frac{m-1}{mn}\)

To combine the fractions on the right side, find a common denominator, which is \(mn\):

\(a = \frac{m}{mn} - \frac{m-1}{mn}\)

\(a = \frac{m - (m-1)}{mn}\)

\(a = \frac{m - m + 1}{mn}\)

\(a = \frac{1}{mn}\)

Calculating the mn-th Term (\(T_{mn}\))

We have found the first term \(a = \frac{1}{mn}\) and the common difference \(d = \frac{1}{mn}\). Now we can find the \(mn\)-th term using the formula \(T_{mn} = a + (mn-1)d\):

\(T_{mn} = \left(\frac{1}{mn}\right) + (mn-1)\left(\frac{1}{mn}\right)\)

\(T_{mn} = \frac{1}{mn} + \frac{mn-1}{mn}\)

Combine the fractions:

\(T_{mn} = \frac{1 + (mn-1)}{mn}\)

\(T_{mn} = \frac{1 + mn - 1}{mn}\)

\(T_{mn} = \frac{mn}{mn}\)

\(T_{mn} = 1\)

Summary of Solution Steps

We used the definition of an arithmetic progression and the given conditions to set up a system of linear equations involving the first term \(a\) and the common difference \(d\). By solving this system, we found \(a = \frac{1}{mn}\) and \(d = \frac{1}{mn}\). Substituting these values into the formula for the \(mn\)-th term, we calculated \(T_{mn}\) to be 1.

Revision Table: Key Concepts in Arithmetic Progression

Additional Information: Properties of AP

Here are some important properties related to Arithmetic Progressions:

• If three numbers \(A, B, C\) are in AP, then \(B\) is the arithmetic mean of \(A\) and \(C\), i.e., \(B = \frac{A+C}{2}\) or \(2B = A+C\).
• In an AP, the sum of terms equidistant from the beginning and end is constant. For example, in an AP with \(N\) terms, \(T_1 + T_N = T_2 + T_{N-1}\), and so on.
• If terms \(T_i, T_j, T_k\) are in AP, then \((i-j)T_k + (j-k)T_i + (k-i)T_j = 0\).
• When selecting terms in an AP, if an odd number of terms are needed, assume them as \(a-d, a, a+d\), etc. If an even number of terms are needed, assume them as \(a-3d, a-d, a+d, a+3d\), etc. (using \(2d\) as common difference for convenience).

These properties can be useful in solving various problems related to arithmetic progressions.