The ratio of the sum of the m and n terms of on 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 given A.P.

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

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

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

But $\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{3}}$ [Given]

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

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

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

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

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

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

$\Rightarrow d = \frac{2 a [n - m]}{[n - m]} = 2 a$ Now,

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

= $\frac{a [1 + 2 m - 2]}{a + [1 + 2 n - n]} = \frac{2 m - 1}{2 n - 1}$

Thus, the ratio of $m^{t h} a n d n^{t h}$ term is (2m + 1) : (2m + 1)

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