The ratio of the sums of $m$ and $n$ terms of an $A.P.$ is $m^{2}: n^{2}$. Show that the ratio of the $m^{th}$ and $n^{th}$ term is $(2m-1):(2n-1)$.

(N/A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$
Given that the ratio of the sum of $m$ terms to the sum of $n$ terms is $\frac{m^2}{n^2}$.
$\frac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \frac{m^2}{n^2}$
$\frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m}{n}$
We need to find the ratio of the $m^{th}$ term to the $n^{th}$ term,which is $\frac{a + (m-1)d}{a + (n-1)d}$.
To get this form,we divide the numerator and denominator of the left side of the equation by $2$:
$\frac{a + \frac{(m-1)}{2}d}{a + \frac{(n-1)}{2}d} = \frac{m}{n}$
Comparing the terms,we substitute $m-1 = 2(m'-1)$ where $m'$ is the term index. Alternatively,replace $m$ with $(2m-1)$ and $n$ with $(2n-1)$ in the ratio $\frac{2a+(m-1)d}{2a+(n-1)d} = \frac{m}{n}$:
$\frac{2a + (2m-1-1)d}{2a + (2n-1-1)d} = \frac{2m-1}{2n-1}$
$\frac{2a + (2m-2)d}{2a + (2n-2)d} = \frac{2m-1}{2n-1}$
$\frac{2[a + (m-1)d]}{2[a + (n-1)d]} = \frac{2m-1}{2n-1}$
$\frac{a + (m-1)d}{a + (n-1)d} = \frac{2m-1}{2n-1}$
Thus,the ratio of the $m^{th}$ term to the $n^{th}$ term is $(2m-1):(2n-1)$.