Let a is the first term and d is the common difference of the AP

Now, S_m = (m/2)*{2a + (m - 1)d}

and S_n = (n/2)*{2a + (n - 1)d}

Now, given

S_m /S_n = m² /n²

=> (m/2)*{2a + (m - 1)d}/(n/2)*{2a + (n - 1)d} = m² /n² 

=> (1/2)*{2a + (m - 1)d}/(1/2)*{2a + (n - 1)d} = m/n

=> {2a + (m - 1)d}/{2a + (n - 1)d} = m/n

=> n{2a + (m - 1)d} = m{2a + (n - 1)d}

=> 2an + n(m - 1)d = 2am + m(n - 1)d

=> 2an - 2am = m(n - 1)d - n(m - 1)d

=> 2a(n - m) = d{m(n - 1) - n(m - 1)}

=> 2a(n - m) = d{mn - m - nm + n}

=> 2a(n - m) = d(n - m)

=> 2a = d

Now, T_m /T_n = {a + (m - 1)d}/{a + (n - 1)d}

=> T_m /T_n = {a + 2a*(m - 1)}/{a + 2a(n - 1)}

=> T_m /T_n = a{1 + 2*(m - 1)}/a*{1 + 2(n - 1)}

=> T_m /T_n = {1 + 2*(m - 1)}/{1 + 2(n - 1)}

=> T_m /T_n = {1 + 2m - 2)}/{1 + 2n - 2}

=> T_m /T_n = {2m - 1)}/{2n - 1}

=> T_m : T_n = {2m - 1)} : {2n - 1}

Hence, the ratio of the mth and nth terms is (2m - 1) : (2n - 1)