To solve this problem, we need to use the formula for the sum of an arithmetic progression (A.P.). Let's break down the problem and solve it step by step.

Given:
The ratio of the sum of the first m terms to the sum of the first n terms of an A.P. is m^2 : n^2.

To prove:
The ratio of any two terms of this A.P. is m : n.

Proof:

1. Let's assume the first term of the A.P. is 'a' and the common difference is 'd'.

2. The sum of the first m terms of an A.P. can be calculated using the formula:

Sn = (m/2)(2a + (m-1)d)

where Sn represents the sum of the first m terms.

3. Similarly, the sum of the first n terms can be calculated using the same formula:

Sn = (n/2)(2a + (n-1)d)

where Sn represents the sum of the first n terms.

4. According to the given condition, the ratio of the sum of the first m terms to the sum of the first n terms is m^2 : n^2. Therefore, we can write:

(m/2)(2a + (m-1)d) / (n/2)(2a + (n-1)d) = m^2 : n^2

5. Simplifying the above expression, we get:

[(2a + (m-1)d) / (2a + (n-1)d)] * [(m/n)^2] = m^2 : n^2

6. Cross multiply the ratio and simplify:

[(2a + (m-1)d) / (2a + (n-1)d)] * (m^2 / n^2) = m^2 : n^2

(2a + (m-1)d) * m^2 = (2a + (n-1)d) * n^2

7. Expanding both sides of the equation, we get:

2am^2 + (m-1)dm^2 = 2an^2 + (n-1)dn^2

8. Rearranging the terms, we have:

(2am^2 - 2an^2) + (m-1)dm^2 - (n-1)dn^2 = 0

9. Factoring out (m-n), we get:

(m-n)(2am + 2an + (m-1)dm + (n-1)dn) = 0

10. Since (m-n) cannot be zero (as m and n are different terms), we can conclude that the second factor must be zero:

2am + 2an + (m-1)dm + (n-1)dn = 0

11. Simplifying the equation further, we get:

2a(m+n) + (m-1)d(m+n) = 0

12. Factoring out (m+n), we have:

(m+n)(2a + (m-1)d) = 0

13. Again, since (m+n) cannot be zero (as m and n are different terms), the first factor must be zero: