Let's assume that the common difference of the arithmetic progression is d.
The sum of the m terms of the A.P. can be calculated using the formula Sn = (m/2)(2a + (m-1)d), where a is the first term.
The sum of the n terms of the A.P. can be calculated using the same formula but with n instead of m.
Given that the ratio of the sum of m and n terms is m^2 : n^2, we can write the following equation:
(m^2/n^2) = [(m/2)(2a + (m-1)d)] / [(n/2)(2a + (n-1)d)]
Simplifying this equation, we get:
(m/n)^2 = (m^2/n^2) * [(2a + (m-1)d) / (2a + (n-1)d)]
(m/n)^2 = (2a + (m-1)d) / (2a + (n-1)d)
Cross multiplying, we get:
(m/n)^2 * (2a + (n-1)d) = 2a + (m-1)d
Expanding, we get:
2am + (m^2 - m)d = 2an + (n^2 - n)d
Rearranging terms, we get:
2am - 2an + md - nd = m^2d - md + n^2d - nd
Grouping like terms, we get:
2a(m - n) + d(m - n) = d(m^2 - m + n^2 - n)
Dividing both sides by (m - n), we get:
2a + d = d(m + n)
Dividing both sides by d, we get:
2a/d + 1 = m + n
We know that the mth term of an A.P. is given by am + d(m-1) and the nth term is given by an + d(n-1).
Dividing these two equations, we get:
(am + d(m-1)) / (an + d(n-1)) = (2a + d) / (2a + d)
Cancelling out the common terms, we get:
am + d(m-1) = an + d(n-1)
Rearranging terms, we get:
am - an + md - nd = d(n-1) - d(m-1)
Grouping like terms, we get:
a(m - n) + d(m - n) = d(n - 1 - m + 1)
Dividing both sides by (m - n), we get:
a + d = d(n - m)
Dividing both sides by d, we get:
a/d + 1 = n - m
Substituting the value of (2a/d + 1) from the previous equation, we get:
n - m = m + n - 2a/d
Cancelling out the common terms, we get:
0 = m - 2a/d
Rearranging terms, we get:
2a/d = m
Therefore, the ratio of the mth term to the nth term is:
2m : 1.