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

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Solution

Let $S_{m}$ and $S_{n}$ be the sum of the first m and first n terms of the AP respectively. Let, a be the first term and d be a common difference
$\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}$

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

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

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

$\Rightarrow 2 a n + m n d - n d + 2 a m + m n d - n d$

$\Rightarrow m d - n d = 2 a m - 2 a n$

$\Rightarrow (m - n) d = 2 a (m - n)$

$\Rightarrow d = 2 a$

Now, the ratio of mth and nth terms is

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

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

$= \frac{2 m - 1}{2 n - 1}$

Thus, ratio of its mth and nth terms is $2 m - 1 : 2 n - 1$

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