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

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Solution

$L e t A b e t h e f i r s t t e r m a n d D b e t h e c o m m o n r a t i o o f a n A . P t h e n,$
$G i v e n$ : $S_{m} = \frac{m}{2} [2 a + (m - 1) d]$
$a n d$ $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Given : $\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}$
$\Rightarrow \frac{\frac{m}{2} [2 a + (m - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{m^{2}}{n^{2}} \Rightarrow \frac{2 a + (m - 1) d}{2 a + (n - 1) d} = \frac{m}{n}$
$\Rightarrow \frac{a + (\frac{m - 1}{2}) d}{a + (\frac{n - 1}{2}) d} = \frac{m}{n} . . . . . . (1)$
$L e t$ $\frac{m - 1}{2} = p - 1 a n d \frac{n - 1}{2} = q - 1,$
$t h e n$ , $m - 1 = 2 p - 2 a n d n - 1 = 2 q - 2$
$⟹ m = 2 p - 1 n = 2 q - 1$
$\Rightarrow \frac{a + (p - 1) d}{a + (q - 1) d} = \frac{2 p - 1}{2 q - 1} . . . . . . (1)$
$\Rightarrow \frac{p^{t h} t e r m}{q^{t h} t e r m} = \frac{2 p - 1}{2 q - 1}$
$H e n c e,$ $\Rightarrow \frac{m^{t h} t e r m}{n^{t h} t e r m} = \frac{2 m - 1}{2 n - 1}$

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