Question

If the sum of first \(m\) terms of an A.P. is \(n\) and the sum of first \(n\) terms of the A.P. is \(m\), then find the sum of \((m+n)\) terms.

Hint:

When given sums of two different terms in an A.P., use simultaneous equations to find \(a\) and \(d\), then compute the required sum.

Answer 1

Step 1: Formula for the sum of \(n\) terms of an A.P.
\[ S_n = \frac{n}{2} [2a + (n - 1)d] \]
Step 2: Write given conditions.
\[ S_m = n = \frac{m}{2} [2a + (m - 1)d] \quad \text{and} \quad S_n = m = \frac{n}{2} [2a + (n - 1)d] \]
Step 3: Simplify both equations.
From (1): \(2n = m[2a + (m - 1)d]\) From (2): \(2m = n[2a + (n - 1)d]\)
Step 4: Eliminate \(a\) and \(d\).
Multiply both sides of (1) and (2) by their respective denominators and simplify. From (1): \[ 2a + (m - 1)d = \frac{2n}{m} \] From (2): \[ 2a + (n - 1)d = \frac{2m}{n} \] Subtract (2) from (1): \[ (m - n)d = \frac{2n}{m} - \frac{2m}{n} \] \[ d = \frac{2(n^2 - m^2)}{mn(m - n)} = \frac{2(n + m)}{mn} \]
Step 5: Find \(a\).
Substitute value of \(d\) in (1): \[ 2a + (m - 1) \frac{2(n + m)}{mn} = \frac{2n}{m} \] Simplify to get \[ a = \frac{n^2 - m^2 + n + m}{mn} \] Step 6: Find \(S_{m+n}\).
\[ S_{m+n} = \frac{m + n}{2} [2a + (m + n - 1)d] \] Substitute \(a\) and \(d\): after simplification, \[ S_{m+n} = 0 \] Step 7: Conclusion.
Hence, the sum of \((m + n)\) terms of the A.P. is \(\boxed{0}\).