If there are n harmonic means between 1 and f | vedclass

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(C) Let $H_1, H_2, \dots, H_n$ be $n$ harmonic means between $a = 1$ and $b = \frac{1}{31}$.Then $1, H_1, H_2, \dots, H_n, \frac{1}{31}$ are in Harmon

Vedclass · Accepted Answer

$14$

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(C) Let $H_1, H_2, \dots, H_n$ be $n$ harmonic means between $a = 1$ and $b = \frac{1}{31}$.Then $1, H_1, H_2, \dots, H_n, \frac{1}{31}$ are in Harmonic Progression $(HP)$.Therefore,$1, \frac{1}{H_1}, \frac{1}{H_2}, \dots, \frac{1}{H_n}, 31$ are in Arithmetic Progression $(AP)$.Let $d$ be the common difference of this $AP$. The $(n+2)^{th}$ term is $31 = 1 + (n+2-1)d$,so $(n+1)d = 30$,which means $d = \frac{30}{n+1}$.The $k^{th}$ harmonic mean $H_k$ is given by $\frac{1}{H_k} = 1 + kd$.Given $\frac{H_7}{H_{n-1}} = \frac{9}{5}$,we have $\frac{1 + (n-1)d}{1 + 7d} = \frac{9}{5}$.Substituting $d = \frac{30}{n+1}$:$5(1 + (n-1)\frac{30}{n+1}) = 9(1 + 7\frac{30}{n+1})$$5(\frac{n+1 + 30n - 30}{n+1}) = 9(\frac{n+1 + 210}{n+1})$$5(31n - 29) = 9(n + 211)$$155n - 145 = 9n + 1899$$146n = 2044$$n = \frac{2044}{146} = 14$.

If there are $n$ harmonic means between $1$ and $\frac{1}{31}$ and the ratio of the $7^{th}$ and $(n - 1)^{th}$ harmonic means is $9:5$,then the value of $n$ is:

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