Understanding the Problem: Sum of AP Terms

The question asks us to find the sum of \((m+n)\) terms of an Arithmetic Progression (AP), given that the sum of \(m\) terms is \(n\) and the sum of \(n\) terms is \(m\). Let the first term of the AP be 'a' and the common difference be 'd'.

The formula for the sum of the first \(k\) terms of an AP is given by:

\( S_k = \frac{k}{2}[2a + (k-1)d] \)

Setting up Equations from Given Information

We are given two conditions:

1. The sum of \(m\) terms is \(n\): \( S_m = n \)
2. The sum of \(n\) terms is \(m\): \( S_n = m \)

Using the sum formula, we can write these conditions as equations:

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

Let's simplify these equations:

• \( m[2a + (m-1)d] = 2n \implies 2am + m(m-1)d = 2n \)
• \( n[2a + (n-1)d] = 2m \implies 2an + n(n-1)d = 2m \)

Solving for 'a' and 'd' or a related expression

We now have a system of two linear equations with two variables, 'a' and 'd'. We can subtract the second equation from the first to eliminate 'a' from some terms and simplify the expressions involving 'd'.

Subtracting Equation 2 from Equation 1:

\( (2am + m(m-1)d) - (2an + n(n-1)d) = 2n - 2m \)

\( 2a(m - n) + [m(m-1) - n(n-1)]d = 2(n - m) \)

\( 2a(m - n) + [m^2 - m - (n^2 - n)]d = -2(m - n) \)

\( 2a(m - n) + [m^2 - n^2 - (m - n)]d = -2(m - n) \)

We can factor the term \( m^2 - n^2 \) as \( (m-n)(m+n) \). Also, factor \( (m-n) \) from the term \( (m-n)d \).

\( 2a(m - n) + [(m-n)(m+n) - (m - n)]d = -2(m - n) \)

Factor out \( (m-n) \) from the terms on the left side:

\( (m - n)[2a + ((m+n) - 1)d] = -2(m - n) \)

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

\( 2a + (m+n - 1)d = -2 \)

This expression \( 2a + (m+n - 1)d \) is part of the formula for the sum of \( (m+n) \) terms.

Calculating the Sum of (m + n) Terms

The sum of \( (m+n) \) terms is given by:

\( S_{m+n} = \frac{m+n}{2}[2a + ((m+n)-1)d] \)

From our previous calculation, we found that \( 2a + (m+n - 1)d = -2 \).

Substitute this value into the formula for \( S_{m+n} \):

\( S_{m+n} = \frac{m+n}{2}(-2) \)

\( S_{m+n} = -(m+n) \)

Conclusion

The sum of \((m+n)\) terms of the AP is \( -(m+n) \).

Revision Table: Key AP Concepts

Here's a quick summary of essential concepts for Arithmetic Progressions:

• Definition: A sequence where the difference between consecutive terms is constant. This constant difference is called the common difference (d).
• General Term (nth term): The nth term of an AP is given by \( a_n = a + (n-1)d \), where 'a' is the first term.
• Sum of First n Terms: The sum of the first n terms is given by \( S_n = \frac{n}{2}[2a + (n-1)d] \) or \( S_n = \frac{n}{2}(a + a_n) \).

Additional Information: Special Cases in AP Sums

While the general formula works for all APs, understanding special cases or properties can be helpful.

• If the common difference \( d = 0 \), the AP is a constant sequence (e.g., 5, 5, 5, ...). The sum of n terms is simply \( n \times a \).
• If the first term \( a = 0 \), the AP is \( 0, d, 2d, 3d, ... \). The sum of n terms becomes \( S_n = \frac{n}{2}[ (n-1)d ] \).
• Problems involving sums of APs often lead to systems of linear equations in 'a' and 'd', as seen in this example. The technique of subtracting equations to simplify or eliminate variables is very common.
• The result \( S_{m+n} = -(m+n) \) is a classic property when \( S_m=n \) and \( S_n=m \) for an AP with \( m \neq n \).