If m times the mth term of an A.P. with non-zero common difference equals n times the nth term of the A.P., where `m ne n`, then (m+n)th term of this A.P. is

Options

A. (m+n) times mth term
B. zero
C. m+n
D. `-(m+n)`

Answer

To solve the problem step by step, we will use the properties of arithmetic progressions (A.P.).

### Step 1: Understand the terms of the A.P.
The nth term of an arithmetic progression (A.P.) is given by the formula:
\[ T_n = a + (n - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write the expressions for the mth and nth terms.
Using the formula for the nth term, we can write:
- The mth term:
\[ T_m = a + (m - 1)d \]
- The nth term:
\[ T_n = a + (n - 1)d \]

### Step 3: Set up the equation based on the given condition.
According to the problem, we have:
\[ m \cdot T_m = n \cdot T_n \]
Substituting the expressions for \( T_m \) and \( T_n \):
\[ m \cdot (a + (m - 1)d) = n \cdot (a + (n - 1)d) \]

### Step 4: Expand and rearrange the equation.
Expanding both sides gives:
\[ ma + m(m - 1)d = na + n(n - 1)d \]
Rearranging this, we get:
\[ ma - na = n(n - 1)d - m(m - 1)d \]
This simplifies to:
\[ (m - n)a = (n(n - 1) - m(m - 1))d \]

### Step 5: Factor the right-hand side.
We can rewrite the right-hand side:
\[ n(n - 1) - m(m - 1) = n^2 - n - (m^2 - m) = n^2 - m^2 - n + m \]
This can be factored as:
\[ (n - m)(n + m - 1) \]

### Step 6: Substitute back into the equation.
Now substituting this back into our equation:
\[ (m - n)a = (n - m)(n + m - 1)d \]
Since \( m \neq n \), we can divide both sides by \( m - n \):
\[ -a = (n + m - 1)d \]
This gives us:
\[ a = -(n + m - 1)d \]

### Step 7: Find the (m+n)th term.
Now we need to find the (m+n)th term:
\[ T_{m+n} = a + (m+n - 1)d \]
Substituting the value of \( a \):
\[ T_{m+n} = -(n + m - 1)d + (m+n - 1)d \]
This simplifies to:
\[ T_{m+n} = -(n + m - 1)d + (m + n - 1)d = 0 \]

### Conclusion
Thus, the (m+n)th term of the A.P. is:
\[ \boxed{0} \]