The ratio of the sum of m and n terms of an AP is $m^{2} : n^{2}$ .show that the ratio of the mth and nth term is (2m-1):(2n-1).

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Solution

Let the first term and common difference of an A.P. is a, d respectively. So, sum of m and n terms of an A.P. is as,
$S_{m} = \frac{m}{2} [2 a + (m - 1) d]$
And,
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
So,
$\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}$
$\Rightarrow \frac{\frac{m}{2} [2 a + (m - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{m^{2}}{n^{2}}$
$\Rightarrow \frac{[2 a + (m - 1) d]}{[2 a + (n - 1) d]} = \frac{m}{n}$
$\Rightarrow n [2 a + (m - 1) d] = m [2 a + (n - 1) d]$
$\Rightarrow 2 a (n - m) = d [m (n - 1) - n (m - 1)]$
$\Rightarrow 2 a (n - m) = d [n - m]$
$\Rightarrow 2 a = d$
Now,
$\frac{T_{m}}{T_{a}} = \frac{a + (m - 1) d}{a + (n - 1) d}$
$= \frac{a + (m - 1) 2 a}{a + (n - 1) 2 a}$
$= \frac{2 m - 1}{2 n - 1}$
$\Rightarrow \frac{T_{m}}{T_{a}} = \frac{2 m - 1}{2 n - 1}$
Hence, ratio of mth and nth term is 2m-1: 2n-1

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