Question

If between 1 and \[\dfrac{1}{{31}}\] there are \[n\] H.M.s and ratio of \[{7^{th}}\] and \[{\left( {n - 1} \right)^{th}}\] harmonic means is 9:5. Then find the value of \[n\].
A. 12
B. 13
C. 14
D. 15

Answer 1

Hint: First we will apply the formula \[{n^{th}}\]term of the harmonic means \[{h_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}\] to calculate the common difference. Then calculate \[{7^{th}}\] and \[{\left( {n - 1} \right)^{th}}\] terms of the harmonic mean. Then equate the ratio of \[{7^{th}}\] and \[{\left( {n - 1} \right)^{th}}\] terms with 9:5 to calculate \[n\].

Formula Used: \[{n^{th}}\]term of the harmonic progress is \[{h_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}\], where \[a\] is the first term and \[d\] is the common difference.

Complete step by step solution:
Given that, there are \[n\] H.M.s in between 1 and \[\dfrac{1}{{31}}\].
The first term of the H.P is 1.
So, \[\dfrac{1}{{31}}\] is \[{\left( {n + 2} \right)^{th}}\] term of the harmonic.
Let \[a\] be the first term of the AP and \[d\] be a common difference.
So, \[a = \dfrac{1}{1} = 1\]
\[{n^{th}}\]term of the harmonic means \[{h_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}\]
The \[{\left( {n + 2} \right)^{th}}\] term of the HM is \[{h_{n + 2}} = \dfrac{1}{{1 + \left( {n + 2 - 1} \right)d}}\].
Since \[{\left( {n + 2} \right)^{th}}\] term of the harmonic is \[\dfrac{1}{{31}}\].
So, \[\dfrac{1}{{31}} = \dfrac{1}{{1 + \left( {n + 2 - 1} \right)d}}\]
\[ \Rightarrow 31 = 1 + \left( {n + 1} \right)d\]
\[ \Rightarrow 31 - 1 = \left( {n + 1} \right)d\]
\[ \Rightarrow d = \dfrac{{30}}{{n + 1}}\]
In the harmonic mean the first and last terms of harmonic progress are not included. So, \[{8^{th}}\] and \[{n^{th}}\] term of HP is \[{7^{th}}\] and \[{\left( {n - 1} \right)^{th}}\] harmonic mean respectively.
Using the formula of \[{n^{th}}\]term of the harmonic progress is \[{h_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}\] to find \[{8^{th}}\] and \[{n^{th}}\] term of HP.
The \[{8^{th}}\] term of HP is \[{h_8} = \dfrac{1}{{1 + \left( {8 - 1} \right) \cdot \dfrac{{30}}{{\left( {n + 1} \right)}}}}\]
                                           \[ = \dfrac{1}{{\dfrac{{n + 1 + 210}}{{\left( {n + 1} \right)}}}}\]
                                         \[ = \dfrac{{n + 1}}{{n + 211}}\]
The \[{n^{th}}\] term of HP is \[{h_n} = \dfrac{1}{{1 + \left( {n - 1} \right) \cdot \dfrac{{30}}{{\left( {n + 1} \right)}}}}\]
                                           \[ = \dfrac{1}{{\dfrac{{n + 1 + 30n - 30}}{{\left( {n + 1} \right)}}}}\]
                                         \[ = \dfrac{{n + 1}}{{31n - 29}}\]
\[{7^{th}}\] and \[{\left( {n - 1} \right)^{th}}\] harmonic means are \[\dfrac{{n + 1}}{{n + 211}}\] and \[\dfrac{{n + 1}}{{31n - 29}}\].
Given that the ratio of \[{7^{th}}\] and \[{\left( {n - 1} \right)^{th}}\] harmonic means is 9:5.
According to the question,
\[\dfrac{{\dfrac{{n + 1}}{{n + 211}}}}{{\dfrac{{n + 1}}{{31n - 29}}}} = \dfrac{9}{5}\]
\[ \Rightarrow \dfrac{{31n - 29}}{{n + 211}} = \dfrac{9}{5}\]
Apply cross multiplication
\[ \Rightarrow 5\left( {31n - 29} \right) = 9\left( {n + 211} \right)\]
\[ \Rightarrow 155n - 145 = 9n + 1899\]
Combine like terms
\[ \Rightarrow 155n - 9n = 1899 + 1455\]
\[ \Rightarrow 146n = 2044\]
\[ \Rightarrow n = \dfrac{{2044}}{{146}}\]
\[ \Rightarrow n = 14\]

Hence option C is correct.

Note: Sometimes students calculate \[{7^{th}}\] and \[{\left( {n - 1} \right)^{th}}\] terms of the harmonic progress to calculate \[{7^{th}}\] and \[{\left( {n - 1} \right)^{th}}\] harmonic mean. In the mean, the first term and last term are not included. \[{8^{th}}\] and \[{n^{th}}\] term of HP is \[{7^{th}}\] and \[{\left( {n - 1} \right)^{th}}\] harmonic mean respectively.