Question

For what value of n, $\dfrac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}$ is the harmonic mean of a and b?

Answer 1

Hint: Think of the basic definition of Harmonic progression, and equate the given expression with $\dfrac{2ab}{a+b}$ to get the answer.

Complete step-by-step answer:
Before starting with the solution to the above question, let us talk about harmonic progression.
In mathematics, harmonic progression is defined as a sequence of numbers, for which the sequence of the reciprocals of its terms gives an arithmetic progression.
We can represent a general harmonic progression as:$\dfrac{1}{a},\dfrac{1}{a+d},\dfrac{1}{a+2d}............$
Where $a,a+d,a+2d...............$ will represent an arithmetic progression.
Now starting with the solution to the given question.
Harmonic mean = $\dfrac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}$
Also, we know that the harmonic mean of two numbers x and y are given by $\dfrac{2xy}{x+y}$ . Therefore, using this in the above equation, we get
$\dfrac{2ab}{a+b}=\dfrac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}$
$\Rightarrow \dfrac{2ab}{a+b}=\dfrac{{{a}^{n+1}}\left( 1+\dfrac{{{b}^{n+1}}}{{{a}^{n+1}}} \right)}{{{a}^{n}}\left( 1+\dfrac{{{b}^{n}}}{{{a}^{n}}} \right)}$
$\Rightarrow \dfrac{2b}{a+b}=\dfrac{\left( 1+\dfrac{{{b}^{n+1}}}{{{a}^{n+1}}} \right)}{\left( 1+\dfrac{{{b}^{n}}}{{{a}^{n}}} \right)}$
\[\Rightarrow \dfrac{2b}{a+b}=\dfrac{\left( 1+{{\left( \dfrac{b}{a} \right)}^{n+1}} \right)}{\left( 1+{{\left( \dfrac{b}{a} \right)}^{n}} \right)}\]
Now we will divide the numerator and denominator of the left-hand side by b. On doing so our expression becomes:
$\dfrac{2}{\dfrac{a}{b}+1}=\dfrac{\left( 1+\dfrac{{{b}^{n+1}}}{{{a}^{n+1}}} \right)}{\left( 1+\dfrac{{{b}^{n}}}{{{a}^{n}}} \right)}$
On letting $\dfrac{b}{a}$ to be k, our expression becomes:
$\dfrac{2}{\dfrac{1}{k}+1}=\dfrac{\left( 1+{{k}^{n+1}} \right)}{\left( 1+{{k}^{n}} \right)}$
$\Rightarrow \dfrac{2k}{k+1}=\dfrac{\left( 1+{{k}^{n+1}} \right)}{\left( 1+{{k}^{n}} \right)}$
Now for further solving the equation, we will cross-multiply. This will give us:
$2k\left( 1+{{k}^{n}} \right)=\left( 1+{{k}^{n+1}} \right)\left( 1+k \right)$
$\Rightarrow 2k+2{{k}^{n+1}}=1+k+{{k}^{n+1}}+{{k}^{n+2}}$
$\Rightarrow {{k}^{n+2}}-{{k}^{n+1}}-k+1=0$
$\Rightarrow {{k}^{n+1}}\left( k-1 \right)-1\left( k-1 \right)=0$
$\Rightarrow \left( {{k}^{n+1}}-1 \right)\left( k-1 \right)=0$
Now we know that a is not equal to b, which means k cannot be equal to 1. Therefore, the only possibility is:
 ${{k}^{n+1}}-1=0$
$\Rightarrow {{k}^{n+1}}=1$
Now, as k is not 1 the power of k should be equal to zero for the equation to be true.
$\therefore n+1=0$
$\Rightarrow n=-1$
Therefore, the answer to the above question is n = -1.

Note: The key to the above question is letting $\dfrac{b}{a}$ to be k. Also, to solve problems like the above one, we need to know the properties and formulas related to different sequences, especially arithmetic and geometric sequence.