Question

Find the value of the following expression if ${{\text{H}}_1},{{\text{H}}_2},...........,{{\text{H}}_{\text{n}}}$ are n harmonic means between a and b (≠a).
$\dfrac{{{{\text{H}}_1} + {\text{a}}}}{{{{\text{H}}_1} - {\text{a}}}} + \dfrac{{{{\text{H}}_{\text{n}}} + {\text{b}}}}{{{{\text{H}}_{\text{n}}} - {\text{b}}}} = $
$
  {\text{A}}{\text{. n + 1}} \\
  {\text{B}}{\text{. n - 1}} \\
  {\text{C}}{\text{. 2n}} \\
  {\text{D}}{\text{. 2n + 3}} \\
$

Answer 1

Hint – In the question we have been given harmonic means, we convert them into arithmetic means. We use the formula for the common difference in arithmetic mean to simplify.

Complete step-by-step answer:
Given Data,
${{\text{H}}_1},{{\text{H}}_2},...........,{{\text{H}}_{\text{n}}}$ are n harmonic means between a and b.

$ \Rightarrow \dfrac{1}{{{{\text{H}}_1}}},\dfrac{1}{{{{\text{H}}_2}}},.........,\dfrac{1}{{{{\text{H}}_{\text{n}}}}}$ are n arithmetic means between $\dfrac{1}{{\text{a}}}$ and $\dfrac{1}{{\text{b}}}$.
(Harmonic mean is the reciprocal of arithmetic mean.)

(Let d be the common difference between the two terms in the harmonic mean)

Now we can write, the common difference d as,

d= $\dfrac{1}{{{{\text{H}}_1}}}{\text{ - }}\dfrac{1}{{\text{a}}}$ -- (1) and d = $\dfrac{1}{{\text{b}}}{\text{ - }}\dfrac{1}{{{{\text{H}}_{\text{n}}}}}$ -- (2)
(The common difference of two consecutive terms is constant.)

=> ${{\text{H}}_1} - {\text{a = - d}}{{\text{H}}_1}{\text{a}}$ and ${{\text{H}}_{\text{n}}} - {\text{b = d}}{{\text{H}}_{\text{n}}}{\text{b}}$

We need,
$\left( {\dfrac{{{{\text{H}}_1} + {\text{a}}}}{{{{\text{H}}_1} - {\text{a}}}}} \right) + \left( {\dfrac{{{{\text{H}}_{\text{n}}} + {\text{b}}}}{{{{\text{H}}_{\text{n}}} - {\text{b}}}}} \right) = \left( {\dfrac{{{{\text{H}}_1} + {\text{a}}}}{{ - {\text{d}}{{\text{H}}_1}{\text{a}}}}} \right) + \left( {\dfrac{{{{\text{H}}_{\text{n}}} + {\text{b}}}}{{{\text{d}}{{\text{H}}_{\text{n}}}{\text{b}}}}} \right)$
$
   \Rightarrow \dfrac{{{{\text{H}}_1}}}{{ - {\text{d}}{{\text{H}}_1}{\text{a}}}} + \dfrac{{\text{a}}}{{ - {\text{d}}{{\text{H}}_1}{\text{a}}}} + \dfrac{{{{\text{H}}_{\text{n}}}}}{{{\text{d}}{{\text{H}}_{\text{n}}}{\text{b}}}} + \dfrac{{\text{b}}}{{{\text{d}}{{\text{H}}_{\text{n}}}{\text{b}}}} \\
   \Rightarrow - \dfrac{1}{{{\text{da}}}} - \dfrac{1}{{{\text{d}}{{\text{H}}_1}}} + \dfrac{1}{{{\text{db}}}} + \dfrac{1}{{{\text{d}}{{\text{H}}_{\text{n}}}}} \\
   \Rightarrow \dfrac{1}{{\text{d}}}\left( {\dfrac{1}{{\text{b}}} - \dfrac{1}{{\text{a}}} + \dfrac{1}{{{{\text{H}}_{\text{n}}}}} - \dfrac{1}{{{{\text{H}}_1}}}} \right){\text{ - - - - - - }}\left( 3 \right) \\
$
From (1) and (2), we get
$\dfrac{1}{{{{\text{H}}_{\text{n}}}}} - \dfrac{1}{{{{\text{H}}_1}}} = \dfrac{1}{{\text{b}}} - \dfrac{1}{{\text{a}}} - 2{\text{d}}$
From arithmetic mean, we know that
$\dfrac{1}{{\text{b}}} - \dfrac{1}{{\text{a}}} = {\text{d}}\left( {{\text{n + 1}}} \right)$

So, $\dfrac{1}{{{{\text{H}}_{\text{n}}}}} - \dfrac{1}{{{{\text{H}}_1}}} = {\text{d}}\left( {{\text{n + 1}}} \right) - 2{\text{d}}$

So Equation (3) becomes $\dfrac{1}{d}(d(n+1) + d(n+1) -2d)$
⟹ (n+1) + (n+1-2)
⟹n+1+n-1
⟹2n.
Hence, $\dfrac{{{{\text{H}}_1} + {\text{a}}}}{{{{\text{H}}_1} - {\text{a}}}} + \dfrac{{{{\text{H}}_{\text{n}}} + {\text{b}}}}{{{{\text{H}}_{\text{n}}} - {\text{b}}}} = $2n
Option C is the correct answer.

Note – In order to solve this type of questions, the key is to convert the harmonic mean into arithmetic mean.
The arithmetic mean, also called the average or average value, is the quantity obtained by summing two or more numbers or variables and then dividing by the number of numbers or variables.
The common difference is the amount between each number in an arithmetic sequence.
d = a (n) - a (n - 1), where a (n) is the last term in the sequence, and a (n - 1) is the previous term in the sequence.