Question

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

Answer 1

Hint: In mathematics, the harmonic mean is one of several kinds of average, and in particular, one of the Pythagorean means. Typically, it is appropriate for situations when the average of rates is desired.The harmonic mean can be expressed as the reciprocal of the arithmetic mean of the reciprocals of the given set of observations.

Complete step by step answer:
The harmonic mean H of the positive real numbers \[{x_1},{x_2},...,{x_n}\] is defined to be \[H = \dfrac{n}{{\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + ... + \dfrac{1}{{{x_n}}}}}\]
There are some special cases as follows:
To find harmonic mean of two numbers :
For the special case of just two numbers \[{x_1}\] and \[{x_2}\] , the harmonic mean can be written as \[H = \dfrac{{2{x_1}{x_2}}}{{{x_1} + {x_2}}}\]
In this special case, the harmonic mean is related to the arithmetic mean \[A = \dfrac{{{x_1} + {x_2}}}{2}\] and the geometric mean \[G = \sqrt {{x_1}{x_2}} \] by \[H = \dfrac{{{G^2}}}{A} = G.\left( {\dfrac{G}{A}} \right)\] .
Since \[\dfrac{G}{A} \leqslant 1\] by the inequality of arithmetic and geometric mean.
Given n HMs between \[1\] and \[\dfrac{1}{{31}}\] . so common difference \[d = \dfrac{{\left( {1 - \dfrac{1}{{31}}} \right)}}{{(n + 1)\left( {\dfrac{1}{{31}}} \right)}}\]
Therefore we get
\[d = \dfrac{{30}}{{(n + 1)}}\]
Now \[\dfrac{1}{{{h_1}}} = 1 + d = \dfrac{{(n + 31)}}{{(n + 1)}}\]
 and \[\dfrac{1}{{{h_7}}} = 1 + 7d = 1 + \dfrac{{7*30}}{{(n + 1)}}\]
which is \[ = \dfrac{{\left( {n + 211} \right)}}{{\left( {n + 1} \right)}}\]
now \[\dfrac{1}{{{h_{n - 1}}}} = 1 + (n - 1)d\]
which becomes \[\dfrac{1}{{{h_{n - 1}}}} = 1 + \dfrac{{(n - 1)*30}}{{(n + 1)}}\]
                                      \[ = \dfrac{{\left( {31n - 29} \right)}}{{\left( {n + 1} \right)}}\]
Therefore \[{h_{n - 1}} = \dfrac{{(n + 1)}}{{(31n - 29)}}\]
And \[{h_7} = \dfrac{{(n + 1)}}{{(n + 211)}}\]
Therefore consider \[\dfrac{{{h_7}}}{{{h_{n - 1}}}} = \dfrac{{(31n - 29)}}{{(n + 211)}}\]
Also we are given that \[\dfrac{{{h_7}}}{{{h_{n - 1}}}} = \dfrac{9}{5}\]
Therefore we get \[\dfrac{9}{5} = \dfrac{{(31n - 29)}}{{(n + 211)}}\]
On cross multiplication we get
\[9(n + 211) = 5(31n - 29)\]
On further simplification we get
\[9n + 1899 = 155n - 145\]
On further simplification we get
\[146n = 2044\]
Therefore we get \[n = \dfrac{{2044}}{{146}}\]
Hence we get \[n = 14\]

So, the correct answer is “Option C”.

Note: The harmonic mean is one of several kinds of average, and in particular, one of the Pythagorean means.The harmonic mean can be expressed as the reciprocal of the arithmetic mean of the reciprocals of the given set of observations. The nth term of the harmonic series is given by \[\dfrac{1}{{a + (n - 1)d}}\] where \[a\] represents the first term and \[d\] represents the common difference.