The ratio of the sums of m and n terms of an A.P. is m² : n², then the ratio of the m^th and n^th terms is 
 

$$\begin{gathered} \frac{{\mathrm{S}}_{\mathrm{m}}}{{\mathrm{S}}_{\mathrm{n}}}=\frac{{\mathrm{m}}^2}{{\mathrm{n}}^2} \\ \frac{{\displaystyle \frac{\mathrm{m}}{2}}\left(2\mathrm{a}+(\mathrm{m}-1)\mathrm{d}\right)}{{\displaystyle \frac{\mathrm{n}}{2}}\left(2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\right)}=\frac{{\mathrm{m}}^2}{{\mathrm{n}}^2} \\ \frac{\mathrm{a}+{\displaystyle \frac{\mathrm{m}-1}{2}}\mathrm{d}}{\mathrm{a}+{\displaystyle \frac{\mathrm{n}-1}{2}}\mathrm{d}}=\frac{\mathrm{m}}{\mathrm{n}} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{m}-1}{2}=\mathrm{m}-1 \Rightarrow \mathrm{m}=2\mathrm{m}-1 \\ \frac{\mathrm{n}-1}{2}=\mathrm{n}-1 \Rightarrow \mathrm{n}=2\mathrm{n}-1 \\ \frac{\mathrm{a}+(\mathrm{m}-1)\mathrm{d}}{\mathrm{a}+(\mathrm{n}-1)\mathrm{d}}=\frac{2\mathrm{m}-1}{2\mathrm{n}-1} \\ \frac{a_m}{a_n}=\frac{2\mathrm{m}-1}{2\mathrm{n}-1} \end{gathered}$$