If the A.M. between mth and nth terms of an A.P. be equal to the A.M. between pth and qth terms of an A.P. then prove that `m+n=p+q`.

Answer

To prove that if the arithmetic mean (A.M.) between the mth and nth terms of an arithmetic progression (A.P.) is equal to the A.M. between the pth and qth terms of an A.P., then \( m + n = p + q \), we can 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 mth term of the A.P. is given by:
\[ a_m = a + (m - 1)d \]
- The nth term of the A.P. is given by:
\[ a_n = a + (n - 1)d \]
- The pth term of the A.P. is given by:
\[ a_p = a + (p - 1)d \]
- The qth term of the A.P. is given by:
\[ a_q = a + (q - 1)d \]

### Step 2: Calculate the A.M. of the mth and nth terms
The A.M. of the mth and nth terms is:
\[ A_m = \frac{a_m + a_n}{2} = \frac{(a + (m - 1)d) + (a + (n - 1)d)}{2} \]
Simplifying this, we get:
\[ A_m = \frac{2a + (m + n - 2)d}{2} = a + \frac{(m + n - 2)d}{2} \]

### Step 3: Calculate the A.M. of the pth and qth terms
The A.M. of the pth and qth terms is:
\[ A_p = \frac{a_p + a_q}{2} = \frac{(a + (p - 1)d) + (a + (q - 1)d)}{2} \]
Simplifying this, we get:
\[ A_p = \frac{2a + (p + q - 2)d}{2} = a + \frac{(p + q - 2)d}{2} \]

### Step 4: Set the two A.M.s equal to each other
According to the problem statement:
\[ A_m = A_p \]
This gives us:
\[ a + \frac{(m + n - 2)d}{2} = a + \frac{(p + q - 2)d}{2} \]

### Step 5: Eliminate \( a \) and simplify
Subtract \( a \) from both sides:
\[ \frac{(m + n - 2)d}{2} = \frac{(p + q - 2)d}{2} \]
Multiplying both sides by 2:
\[ (m + n - 2)d = (p + q - 2)d \]

### Step 6: Divide by \( d \) (assuming \( d \neq 0 \))
Assuming \( d \neq 0 \), we can divide both sides by \( d \):
\[ m + n - 2 = p + q - 2 \]

### Step 7: Rearranging the equation
Rearranging gives us:
\[ m + n = p + q \]

### Conclusion
Thus, we have proved that if the A.M. between the mth and nth terms of an A.P. is equal to the A.M. between the pth and qth terms of an A.P., then:
\[ m + n = p + q \]
Hence proved.

---