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The nth term of $H . P$ , whose first two terms are 6 and 3 respectively is

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\frac{6}{n}

6 n

\frac{n}{6}

\frac{n}{3}

answer is A.

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Detailed Solution

a_{1} = 6, a_{2} = 3

H . P

$a_{1} = \frac{1}{6}, a_{2} = \frac{1}{3}$ in $A . P$

$d = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$

$a_{n} = a + (n - 1) d$

$= \frac{1}{6} + (n - 1) \frac{1}{6}$

$a_{n} = \frac{n}{6}$ in $A . P$

$a_{n} = \frac{6}{n}$ in $H . P$

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