If the mth term of an AP is $\frac{1}{n}$ and the nth term is $\frac{1}{m}$ , show that the sum of mn terms is $\frac{1}{2} (m n + 1)$ .

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Solution

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

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

Subtracting equation 2 form equation 1, we get,
$(m - n) d = \frac{1}{n} - \frac{1}{m}$

$(m - n) d = \frac{m - n}{m n}$

$d = \frac{1}{m n}$

Putting this value of d in equation 1, we get,
$a + (m - 1) \frac{1}{m n} = \frac{1}{n}$

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

$a = \frac{1}{m n}$

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

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

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

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If the mth term of an AP is 1n and the nth term is 1m, show that the sum of mn terms is 12(mn+1).

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If the mth term of an AP is $\frac{1}{n}$ and the nth term is $\frac{1}{m}$ , show that the sum of mn terms is $\frac{1}{2} (m n + 1)$ .