Harmonic Mean

The harmonic mean of $n$ numbers is defined as $n$ divided by the sum of the reciprocals of these numbers. The harmonic mean of the numbers $x_1, x_2, x_3 \cdots x_n $ is:

\[ \displaystyle\frac{n}{\displaystyle \frac{1}{x_1} + \displaystyle\frac{1}{x_2} +\displaystyle \frac{1}{x_3} + …+ \displaystyle\frac{1}{x_n} } \]

The harmonic mean of the numbers $x_1 $ and $ x_2 $ is:

\[ \displaystyle\frac{2}{\displaystyle\frac{1}{x_1} + \displaystyle\frac{1}{x_2}} =\displaystyle \frac{2 x_1 x_2}{x_1 + x_2} \]

Note:

Let $A$ be the arithmetic mean, $G$ be the geometric mean, and $H$ be the harmonic mean of two numbers, then:

\[ G^2 = A \cdot H \] and \[ A >G < H \]

Question 1

If the arithmetic mean and the harmonic mean of a negative number and its reciprocal are equal, what is the value of this number?

\[
\text{A)} -1 \quad
\text{B) }-2 \quad
\text{C) } -3 \quad
\text{D) } -4 \quad
\text{E) } -5
\]

Solution:

If we let this number be $ x $, its reciprocal becomes $ \frac{1}{x} $. According to this,

\[ \frac{x + \frac{1}{x} }{2} = \frac{2 \cdot x \cdot \frac{1}{x} }{x + \frac{1}{x} } \Rightarrow (x + \frac{1}{x} )^2 = 4 \]

\[ \left| x + \frac{1}{x} \right| = 2 \]

Since $ x < 0 $, it follows that $ x + \frac{1}{x} < 0 $. From here,

\[ \Rightarrow x + \frac{1}{x} = -2 \]

\[ \Rightarrow \frac{x^2+1 }{x} = -2 \]

\[ \Rightarrow x^2 + 2x + 1 = 0\]

\[ \Rightarrow x = -1 \]

$\textbf{Answer: A} $

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