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

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Solution

Let a be the first term and d the common difference of the given A.P. Then, the sums of m and n terms are given by

$S_{m}$ = ( $\frac{m}{2}$ ) [2a + (m – 1) d], and

$S_{n}$ = ( $\frac{n}{2}$ ) [2a + (n – 1) d]

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

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

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

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

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

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

$\Rightarrow d = 2 a$

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

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