Let $a$ and $b$ be the first term and the common difference of the $A.P.$ respectively. According to the given condition,

$\frac{{{\rm{ Sum}}\,\,{\rm{of }}\,\,m\,\,{\rm{ terms }}}}{{{\rm{ Sum }}\,\,{\rm{of}}\,{\rm{ }}n{\rm{ }}\,\,{\rm{terms }}}} = \frac{{{m^2}}}{{{n^2}}}$

$\Rightarrow \frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) d]}=\frac{m^{2}}{n^{2}}$

$\Rightarrow \frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n}$        ……..$(1)$

Putting $m=2 m-1$ and $n=2 n-1,$ we obtain

$\frac{2 a+(2 m-2) d}{2 a+(2 n-2) d}=\frac{2 m-1}{2 n-1}$

$\Rightarrow \frac{a+(m-1) d}{a+(n-1) d}=\frac{2 m-1}{2 n-1}$         ……….$(2)$

$\frac{{{m^{th}}\,\,{\rm{ term}}\,\,{\rm{ of}}\,\,{\rm{ A}}{\rm{.P}}{\rm{. }}}}{{{n^{{\rm{th }}}}\,\,{\rm{ term }}\,\,{\rm{of}}\,\,{\rm{ A}}{\rm{.P}}{\rm{. }}}} = \frac{{a + (m – 1)d}}{{a + (n – 1)d}}$

From $(2)$ and $(3),$ we obtain

$\frac{m^{H h} \text { termof A.P. }}{n^{t h} \text { termof A.P. }}=\frac{2 m-1}{2 n-1}$

Thus, the given result is proved.