Find the 9^{th} term of the H.P. 6, 4, 3, ldo | vedclass

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(B) The given sequence is $6, 4, 3, \ldots$ which is a $H.P.$The sequence of reciprocals of its terms is $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \ldot

Vedclass · Accepted Answer

$\frac{6}{5}$

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(B) The given sequence is $6, 4, 3, \ldots$ which is a $H.P.$The sequence of reciprocals of its terms is $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \ldots$ which is an $A.P.$Here,the first term $a = \frac{1}{6}$ and the common difference $d = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.For the $9^{th}$ term of the $A.P.$:$a_9 = \frac{1}{6} + (9-1) 	imes \frac{1}{12}$$a_9 = \frac{1}{6} + 8 	imes \frac{1}{12} = \frac{1}{6} + \frac{8}{12} = \frac{1}{6} + \frac{2}{3} = \frac{1+4}{6} = \frac{5}{6}$.Since the $H.P.$ is the reciprocal of the $A.P.$,the $9^{th}$ term of the $H.P.$ is the reciprocal of $a_9$.Therefore,the $9^{th}$ term of the $H.P.$ is $\frac{6}{5}$.

Find the $9^{th}$ term of the $H.P. 6, 4, 3, \ldots$

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