The fifth term of the H.P.,2, 2frac{1}{2}, 3f | vedclass

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(D) The given series $2, 2\frac{1}{2}, 3\frac{1}{3}, \dots$ is in $H.P.$Taking the reciprocals,the series $\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \do

Vedclass · Accepted Answer

$10$

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(D) The given series $2, 2\frac{1}{2}, 3\frac{1}{3}, \dots$ is in $H.P.$Taking the reciprocals,the series $\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \dots$ is in $A.P.$Here,the first term $a = \frac{1}{2}$.The common difference $d = \frac{2}{5} - \frac{1}{2} = \frac{4-5}{10} = -\frac{1}{10}$.The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$.For the $5^{th}$ term of the $A.P.$: $T_5 = \frac{1}{2} + (5-1)(-\frac{1}{10}) = \frac{1}{2} - \frac{4}{10} = \frac{5-4}{10} = \frac{1}{10}$.Since the $H.P.$ is the reciprocal of the $A.P.$,the $5^{th}$ term of the $H.P.$ is the reciprocal of $\frac{1}{10}$,which is $10$.

The fifth term of the $H.P.$,$2, 2\frac{1}{2}, 3\frac{1}{3}, \dots$ will be

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