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

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(C) Let the $n$ harmonic means be $H_1, H_2, \dots, H_n$ between $1$ and $\frac{1}{31}$.Then $1, H_1, H_2, \dots, H_n, \frac{1}{31}$ are in Harmonic P

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$14$

Vedclass · Answer

(C) Let the $n$ harmonic means be $H_1, H_2, \dots, H_n$ between $1$ and $\frac{1}{31}$.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 the common difference of this $AP$ be $d$. The first term $a = 1$ and the $(n+2)^{th}$ term is $31$.$a + (n + 2 - 1)d = 31 \implies 1 + (n + 1)d = 31 \implies (n + 1)d = 30 \implies d = \frac{30}{n + 1}$.The $k^{th}$ harmonic mean $H_k$ is given by $\frac{1}{a + kd}$.Given $\frac{H_7}{H_{n-1}} = \frac{9}{5} \implies \frac{a + (n - 1)d}{a + 7d} = \frac{9}{5}$.Substituting $a = 1$: $\frac{1 + (n - 1)d}{1 + 7d} = \frac{9}{5} \implies 5 + 5(n - 1)d = 9 + 63d$.$5(n - 1)d - 63d = 4 \implies d(5n - 5 - 63) = 4 \implies d(5n - 68) = 4$.Substitute $d = \frac{30}{n + 1}$: $\frac{30(5n - 68)}{n + 1} = 4 \implies 15(5n - 68) = 2(n + 1)$.$75n - 1020 = 2n + 2 \implies 73n = 1022 \implies n = 14$.

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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