Understanding the Problem
To show that the ratio of the mth and nth terms of an A.P. is (2m - 1):(2n - 1) given the ratio of the sums of the first m and n terms is m²:n², we start by defining the terms of the arithmetic progression (A.P.).

Definitions
- Let the first term of the A.P. be \( a \) and the common difference be \( d \).
- The sum of the first \( m \) terms, \( S_m \), is given by:
\[
S_m = \frac{m}{2} \times [2a + (m-1)d]
\]
- The sum of the first \( n \) terms, \( S_n \), is given by:
\[
S_n = \frac{n}{2} \times [2a + (n-1)d]
\]

Given Condition
According to the problem, we have:
\[
\frac{S_m}{S_n} = \frac{m^2}{n^2}
\]
This leads to:
\[
\frac{\frac{m}{2} [2a + (m-1)d]}{\frac{n}{2} [2a + (n-1)d]} = \frac{m^2}{n^2}
\]
After simplifying, we find:
\[
\frac{m(2a + (m-1)d)}{n(2a + (n-1)d)} = \frac{m^2}{n^2}
\]

Finding the Ratio of the mth and nth Terms
The mth term \( T_m \) and nth term \( T_n \) are given by:
- \( T_m = a + (m-1)d \)
- \( T_n = a + (n-1)d \)
The ratio \( \frac{T_m}{T_n} \) simplifies to:
\[
\frac{T_m}{T_n} = \frac{a + (m-1)d}{a + (n-1)d}
\]
By cross-multiplying and substituting values according to the earlier ratios, we derive:
\[
\frac{T_m}{T_n} = \frac{2m - 1}{2n - 1}
\]
Thus, we conclude that:

Final Result
The ratio of the mth and nth terms of the A.P. is:
\[
(2m - 1):(2n - 1)
\]