If m times the mth term of an AP is equal to n times its nth term, show that $(m + n) t h$ term is zero.

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Solution

Let $a$ be the first term and $d$ be the common difference of the given A.P.
Also, let $t_{m}$ be the $(m)^{t h}$ term and $t_{n}$ be the $(n)^{t h}$ term of the progression.
Given that,
$m \times t_{m} = n \times t_{n} \Rightarrow m t_{m} = n t_{n}$
For an A.P., we know that, $a_{n} = a + (n - 1) d$
$∴ m t_{m} = n t_{n} \Rightarrow m [a + (m - 1) d] = n [a + (n - 1) d]$
$\Rightarrow m [a + (m - 1) d] - n [a + (n - 1) d] = 0$
$\Rightarrow a (m - n) + [m (m - 1) - n (n - 1)] d = 0$
$\Rightarrow a (m - n) + [(m^{2} - n^{2}) - (m - n)] d = 0$
$\Rightarrow a (m - n) + (m - n) (m + n - 1) d = 0$
$\Rightarrow (m - n) [a + (m + n - 1) d] = 0$
$\Rightarrow a + ((m + n) - 1) d = 0$
$∴ a_{m + n} = 0$

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