32. Show that the sum of `(m+n)^(th)` and `(m-n)^(th)` terms of an A.P. is equal to twice the `m^(th)` term

Explanation

To show that the sum of the \((m+n)^{th}\) and \((m-n)^{th}\) terms of an Arithmetic Progression (A.P.) is equal to twice the \(m^{th}\) term, we will follow these steps:

### Step 1: Define the terms of the A.P.
Let the first term of the A.P. be \(a\) and the common difference be \(d\). The \(k^{th}\) term of an A.P. is given by the formula:

\[ T_k = a + (k-1)d \]

### Step 2: Write the expressions for the required terms
We need to find the \((m+n)^{th}\) term and the \((m-n)^{th}\) term.

- The \((m+n)^{th}\) term is:
\[ T_{m+n} = a + (m+n-1)d \]

- The \((m-n)^{th}\) term is:
\[ T_{m-n} = a + (m-n-1)d \]

### Step 3: Calculate the sum of the two terms
Now, we will calculate the sum \(T_{m+n} + T_{m-n}\):

\[ T_{m+n} + T_{m-n} = \left(a + (m+n-1)d\right) + \left(a + (m-n-1)d\right) \]

Combining the terms, we get:

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

### Step 4: Simplify the expression
Now, simplify the expression inside the parentheses:

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

Thus, we can rewrite the sum as:

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

Factoring out the 2, we have:

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

### Step 5: Relate it to the \(m^{th}\) term
Now, we can express this in terms of the \(m^{th}\) term \(T_m\):

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

So, multiplying \(T_m\) by 2 gives:

\[ 2T_m = 2\left(a + (m-1)d\right) = 2a + 2(m-1)d \]

### Conclusion
We have shown that:

\[ T_{m+n} + T_{m-n} = 2T_m \]

Thus, the sum of the \((m+n)^{th}\) and \((m-n)^{th}\) terms of an A.P. is equal to twice the \(m^{th}\) term.