Let `t_r` denote the `r^(th)` term of an A.P. Also suppose `t_m=1/n` and `t_n=1/m` for some positive integers `m` and `n` then which of the following is necessarily a root of the equation? `(l+m-2n)x^2+(m+n-2l)x+(n+l-2m)=0`

Options

A. `t_(n)`
B. `t_(m)`
C. `t_(m +n)`
D. `t_(m)`

Answer

To solve the problem step by step, we will analyze the information given and derive the necessary conclusions.

### Step 1: Understand the terms of the A.P.
Let \( t_r \) denote the \( r^{th} \) term of an arithmetic progression (A.P.). The general formula for the \( r^{th} \) term of an A.P. is given by:
\[ t_r = a + (r - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Set up the equations based on the given information.
We are given:
- \( t_m = \frac{1}{n} \)
- \( t_n = \frac{1}{m} \)

Using the formula for the \( r^{th} \) term, we can write:
1. For \( t_m \):
\[ a + (m - 1)d = \frac{1}{n} \quad \text{(Equation 1)} \]
2. For \( t_n \):
\[ a + (n - 1)d = \frac{1}{m} \quad \text{(Equation 2)} \]

### Step 3: Subtract Equation 1 from Equation 2.
Subtracting Equation 1 from Equation 2 gives:
\[ [a + (n - 1)d] - [a + (m - 1)d] = \frac{1}{m} - \frac{1}{n} \]
This simplifies to:
\[ (n - 1)d - (m - 1)d = \frac{1}{m} - \frac{1}{n} \]
\[ (n - m)d = \frac{n - m}{mn} \]

### Step 4: Solve for \( d \).
Assuming \( n \neq m \) (since \( m \) and \( n \) are positive integers), we can divide both sides by \( n - m \):
\[ d = \frac{1}{mn} \]

### Step 5: Substitute \( d \) back into Equation 1 to find \( a \).
Substituting \( d \) into Equation 1:
\[ a + (m - 1)\left(\frac{1}{mn}\right) = \frac{1}{n} \]
This simplifies to:
\[ a + \frac{m - 1}{mn} = \frac{1}{n} \]
Rearranging gives:
\[ a = \frac{1}{n} - \frac{m - 1}{mn} \]
\[ a = \frac{m - 1}{mn} + \frac{1}{n} - \frac{m - 1}{mn} \]
\[ a = \frac{1}{n} - \frac{m - 1}{mn} = \frac{1}{n} - \frac{m - 1}{mn} \]

### Step 6: Check if \( x = 1 \) is a root of the given quadratic equation.
The quadratic equation is:
\[ (l + m - 2n)x^2 + (m + n - 2l)x + (n + l - 2m) = 0 \]
Substituting \( x = 1 \):
\[ (l + m - 2n) + (m + n - 2l) + (n + l - 2m) = 0 \]
Simplifying:
\[ l + m - 2n + m + n - 2l + n + l - 2m = 0 \]
Combining like terms:
\[ 0 = 0 \]
This confirms that \( x = 1 \) is indeed a root.

### Conclusion
Thus, we conclude that the necessary root of the equation is \( x = 1 \).

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