The sum of the ratio of m and n th term is (m)2:(n)2 to show the ratio of mn th term is (2m-1):(2n-1)

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Solution

Let $S_{m}$ and $S_{n}$ the sum of the m and n terms of the A.P. respectively
$S_{m} = \frac{m}{2} [2 a + (m - 1) d] S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Here, a is the first terms and d is the common difference.
$\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}$
$∴ \frac{\frac{m}{2} [2 a + (m - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{m^{2}}{n^{2}} \Rightarrow \frac{m [2 a + (m - 1) d]}{n [2 a + (n - 1) d]} = \frac{m^{2}}{n^{2}} \Rightarrow \frac{2 a + (m - 1) d}{2 a + (n - 1) d} = \frac{m}{n}$
$\Rightarrow [2 a + (m - 1) d] n = [2 a + (n - 1) d] m \Rightarrow 2 a n + (m - 1) d n = 2 a m + (n - 1) d m \Rightarrow d [(m - 1) n - (n - 1) m] = 2 a m - 2 a n \Rightarrow d (m n - n - n m + m) = 2 a (m - n) \Rightarrow d (m - n) = 2 a (m - n) \Rightarrow d = 2 a$
Now,
$t_{m} = a + (m - 1) d t_{n} = a + (n - 1) d \frac{t_{m}}{t_{n}} = \frac{a + (m - 1) 2 a}{a + (n - 1) 2 a} = \frac{a [1 + 2 (m - 1)]}{a [1 + (n - 1)]} = \frac{1 + 2 m - 2}{1 + 2 n - 2} = \frac{2 m - 1}{2 n - 1}$
Thus, $m^{t h}$ term: $n^{t h}$

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