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 $n^{t h}$ term is (2m - 1) : (2n - 1).

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Solution

Let a be the first term and d be the common difference of the given A.P. Then, the sum of m and n terms are given by
$S_{m} = \frac{m}{2} [2 a + (m - 1) d]$ , and $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

Then,
$\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}$

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

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

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

Therefore,

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

Hence proved

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