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 terms is (2m - 1) : (2n - 1).

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Solution

Let a and d be the first term and common difference of the given AP

Then, the sum of m and n terms are given by $s_{m} = \frac{m}{2} [2 a + (m - 1) d]$

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

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

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

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

Now, the ratio of the mth and nth terms, $\frac{a_{m}}{a_{n}} = \frac{a + (m - 1) d}{a + (n - 1) d} = \frac{a + (m - 1) 2 a}{a + (n - 1) 2 a}$ [using Eq. (i)]

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

Hence proved.

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