Given $\frac{1}{f} = (n - 1) (\frac{1}{a} + \frac{1}{b})$ , make 'n' as the subject of the formula. Hence, find 'n' if $f = 1.2, a = 2$ and $b = 3$ .

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Solution

Given, $\frac{1}{f} = (n - 1) (\frac{1}{a} + \frac{1}{b})$
$=> \frac{1}{f} = (n - 1) (\frac{a + b}{a b})$
$=> \frac{a b}{f (a + b)} = (n - 1)$
$=> n = \frac{a b}{f (a + b)} + 1$
When $f = 1.2, a = 2, b = 3$
$n = \frac{2 \times 3}{1.2 \times (2 + 3)} + 1 = \frac{6}{1.2 \times 5} + 1 = 2$

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