Understanding Arithmetic Progressions (AP)

An arithmetic progression (AP) is 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'. The first term of an AP is usually denoted by 'a'.

The formula for the r-th term (\(t_r\)) of an arithmetic progression is given by:

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

where:

• \(t_r\) is the r-th term
• \(a\) is the first term
• \(r\) is the term number
• \(d\) is the common difference

Setting Up Equations for the AP Problem

We are given information about two specific terms in the arithmetic progression:

• The m-th term, \(t_m\), is equal to \(\dfrac{1}{n}\).
• The n-th term, \(t_n\), is equal to \(\dfrac{1}{m}\).

Using the formula for the r-th term, we can write these facts as equations:

1. \(t_m = a + (m-1)d = \dfrac{1}{n}\)
2. \(t_n = a + (n-1)d = \dfrac{1}{m}\)

We now have a system of two linear equations with two variables, 'a' (the first term) and 'd' (the common difference).

Solving for the First Term 'a' and Common Difference 'd'

To find 'a' and 'd', we can solve this system of equations. A common method is subtraction to eliminate 'a'.

Subtract Equation 2 from Equation 1:

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

Simplify the left side:

\(a + md - d - a - nd + d = (m-n)d\)

Simplify the right side by finding a common denominator (mn):

\(\dfrac{1}{n} - \dfrac{1}{m} = \dfrac{m}{mn} - \dfrac{n}{mn} = \dfrac{m-n}{mn}\)

So, the equation becomes:

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

Assuming \(m \neq n\), we can divide both sides by \((m-n)\) to find 'd':

\(d = \dfrac{\dfrac{m-n}{mn}}{m-n} = \dfrac{m-n}{mn(m-n)} = \dfrac{1}{mn}\)

Now that we have the common difference \(d = \dfrac{1}{mn}\), we can substitute this value back into either Equation 1 or Equation 2 to find 'a'. Let's use Equation 1:

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

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

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

Now, isolate 'a':

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

To subtract the fractions, find a common denominator, which is mn:

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

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

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

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

So, we have found the first term \(a = \dfrac{1}{mn}\) and the common difference \(d = \dfrac{1}{mn}\).

Calculating the mn-th Term (t_mn)

We need to find the mn-th term of the arithmetic progression, which is \(t_{mn}\). Using the formula \(t_r = a + (r-1)d\), we substitute \(r = mn\), \(a = \dfrac{1}{mn}\), and \(d = \dfrac{1}{mn}\):

\(t_{mn} = a + (mn-1)d\)

\(t_{mn} = \dfrac{1}{mn} + (mn-1)\left(\dfrac{1}{mn}\right)\)

\(t_{mn} = \dfrac{1}{mn} + \dfrac{mn-1}{mn}\)

Since the denominators are the same, we can add the numerators:

\(t_{mn} = \dfrac{1 + (mn-1)}{mn}\)

\(t_{mn} = \dfrac{1 + mn - 1}{mn}\)

\(t_{mn} = \dfrac{mn}{mn}\)

\(t_{mn} = 1\)

The mn-th term of the arithmetic progression is 1.

Checking the Options

Let's look at the given options to find which one matches our result for \(t_{mn}\):

1. \(\dfrac{1}{mn}\)
2. 4
3. \(\dfrac{1}{m} + \dfrac{1}{n}\)
4. 1

Our calculated value for \(t_{mn}\) is 1, which corresponds to Option 4.

Revision Table: Key Arithmetic Progression Concepts

Additional Information: Properties of AP

Here are some additional points about arithmetic progressions:

• If three numbers a, b, and c are in AP, then \(2b = a + c\). 'b' is called the arithmetic mean of 'a' and 'c'.
• If terms \(t_p, t_q, t_r\) of an AP are in AP, then p, q, r are also in AP.
• If \(t_m = \dfrac{1}{n}\) and \(t_n = \dfrac{1}{m}\) in an AP, it's a classic problem structure. The common difference \(d\) and the first term \(a\) are often found to be related to mn. In this specific case, both turned out to be \(\dfrac{1}{mn}\).
• The mn-th term \(t_{mn}\) in this specific scenario (\(t_m = \dfrac{1}{n}\) and \(t_n = \dfrac{1}{m}\)) is always 1, provided \(m \neq n\). This is a noteworthy property that can sometimes help in quickly solving similar problems or verifying results.