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

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Solution

Step $1 :$ Simplification of ratio
Ratio of the sums of $m$ and $n$ terms of an A.P. is $\frac{m^{2}}{n^{2}}$
i.e. $\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}$
$\Rightarrow \frac{\frac{m}{2} [2 a + (m - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{m^{2}}{n^{2}}$
$\Rightarrow \frac{2 a + (m - 1) d}{2 a + (n - 1) d} = \frac{m}{n} \dots (i)$

Step $2 :$ Finding ratio of $m^{t h}$ and $n^{t h}$ terms.
Substituting $m = 2 m - 1$ and $n = 2 n - 1$ in equation $(i),$ we get
$\frac{[2 a + (2 m - 2) d]}{[2 a + (2 n - 2) d]} = \frac{2 m - 1}{2 n - 1}$
$\frac{[a + (m - 1) d]}{[a + (n - 1) d]} = \frac{2 m - 1}{2 n - 1}$
$∴ \frac{a_{m}}{a_{n}} = \frac{2 m - 1}{2 n - 1}$ Hence approved

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