Question

If \[\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}\] is the geometric mean of a and b then what is the value of \[n\] ? \[\left( {a,b \in {R^ + }} \right)\]
A.0
B.1
C. \[\dfrac{1}{2}\]
D.-2

Answer 1

Hint: To solve the question we use the geometric mean formula i.e., Geometric mean of two numbers a and b is given by \[\sqrt {ab} \], and by equating the given geometric mean i.e., \[\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}\] to the formula i.e., \[\sqrt {ab} \] and after that solve the equation using exponents formulas we will get the desired value of n.

Formula used: We will use the formula of geometric mean of two numbers a and b is given by \[\sqrt {ab} \], And we will also use the exponents formulas i.e., \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\], \[\dfrac{1}{{{a^m}}} = {a^{ - m}}\], \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\] and \[{a^0} = 1\].

Complete Step-by-Step Solution:
Given that If \[\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}\] is the geometric mean of a and b,
We know that the geometric mean is the positive square root of the product of two numbers i.e, the geometric mean of \[a\] and \[b\] will be \[\sqrt {ab} \],
From the given data,
 \[ \Rightarrow \dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}} = \sqrt {ab} \]
By cross multiplying we get,
Now we will simplify, then we will get,
 \[ \Rightarrow \dfrac{{{a^n} + {b^n}}}{{\sqrt {ab} }} = {a^{n - 1}} + {b^{n - 1}}\]
Now we will separate and divide each term, then we will get,
 \[ \Rightarrow \dfrac{{{a^n}}}{{{a^{\dfrac{1}{2}}} \cdot {b^{\dfrac{1}{2}}}}} + \dfrac{{{b^n}}}{{{a^{\dfrac{1}{2}}} \cdot {b^{\dfrac{1}{2}}}}} = {a^{n - 1}} + {b^{n - 1}}\]
Now again we will simplify by using formula, \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\] ,
 \[ \Rightarrow \dfrac{{{a^{n - \dfrac{1}{2}}}}}{{{b^{\dfrac{1}{2}}}}} + \dfrac{{{b^{n - \dfrac{1}{2}}}}}{{{a^{\dfrac{1}{2}}}}} = {a^{n - 1}} + {b^{n - 1}}\]
Again we will simplify by using formula, \[\dfrac{1}{{{a^m}}} = {a^{ - m}}\] ,
 \[ \Rightarrow {a^{n - \dfrac{1}{2}}} \cdot {b^{\dfrac{{ - 1}}{2}}} + {b^{n - \dfrac{1}{2}}} \cdot {a^{\dfrac{{ - 1}}{2}}} = {a^{n - 1}} + {b^{n - 1}}\]
Further we will simplify then we will get,
 \[ \Rightarrow {a^{n - \dfrac{1}{2}}} \cdot {b^{\dfrac{{ - 1}}{2}}} + {b^{n - \dfrac{1}{2}}} \cdot {a^{\dfrac{{ - 1}}{2}}} - {a^{n - 1}} - {b^{n - 1}} = 0\]
Now we will take all like terms,
 \[ \Rightarrow {a^{n - \dfrac{1}{2}}} \cdot {b^{\dfrac{{ - 1}}{2}}} - {a^{n - 1}} + {b^{n - \dfrac{1}{2}}} \cdot {a^{\dfrac{{ - 1}}{2}}} - {b^{n - 1}} = 0\]
Now we will simplify, then we will get,
 \[ \Rightarrow {a^{n - \dfrac{1}{2}}}\left( {{b^{\dfrac{{ - 1}}{2}}} - {a^{\dfrac{{ - 1}}{2}}}} \right) + {b^{n - \dfrac{1}{2}}}\left( {{a^{\dfrac{{ - 1}}{2}}} - {b^{\dfrac{{ - 1}}{2}}}} \right) = 0\]
 \[ \Rightarrow {a^{n - \dfrac{1}{2}}}\left( {{b^{\dfrac{{ - 1}}{2}}} - {a^{\dfrac{{ - 1}}{2}}}} \right) - {b^{n - \dfrac{1}{2}}}\left( {{b^{\dfrac{{ - 1}}{2}}} - {a^{\dfrac{{ - 1}}{2}}}} \right) = 0\]
Now we will take out the like terms, then we will get,
 \[ \Rightarrow \left( {{a^{n - \dfrac{1}{2}}} - {b^{n - \dfrac{1}{2}}}} \right)\left( {{a^{\dfrac{{ - 1}}{2}}} - {b^{\dfrac{{ - 1}}{2}}}} \right) = 0\]
Now we will equate each term to 0,
 \[ \Rightarrow \left( {{a^{n - \dfrac{1}{2}}} - {b^{n - \dfrac{1}{2}}}} \right) = 0\] and \[ \Rightarrow \left( {{a^{\dfrac{{ - 1}}{2}}} - {b^{\dfrac{{ - 1}}{2}}}} \right) = 0\]
Now we will solve the equations, take 2nd equation
 \[ \Rightarrow \left( {{a^{\dfrac{{ - 1}}{2}}} - {b^{\dfrac{{ - 1}}{2}}}} \right) = 0\]
 \[ \Rightarrow {a^{\dfrac{{ - 1}}{2}}} = {b^{\dfrac{{ - 1}}{2}}}\]
Now we will take 1st equation,
 \[ \Rightarrow \left( {{a^{n - \dfrac{1}{2}}} - {b^{n - \dfrac{1}{2}}}} \right) = 0\]
 \[ \Rightarrow {a^{n - \dfrac{1}{2}}} - {b^{n - \dfrac{1}{2}}} = 0\]
Now we will simplify, then we will get,
 \[ \Rightarrow {a^{n - \dfrac{1}{2}}} = {b^{n - \dfrac{1}{2}}}\]
Now we will take all terms to side,
 \[ \Rightarrow \dfrac{{{a^{n - \dfrac{1}{2}}}}}{{{b^{n - \dfrac{1}{2}}}}} = 1\]
Now we will use formula, \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\], we get,
 \[ \Rightarrow \left( {\dfrac{{{a^{n - \dfrac{1}{2}}}}}{{{b^{n - \dfrac{1}{2}}}}}} \right) = 1\]
Now we will use the formula, \[{a^0} = 1\],
 \[ \Rightarrow {\left( {\dfrac{a}{b}} \right)^{n - \dfrac{1}{2}}} = {\left( {\dfrac{a}{b}} \right)^0}\]
Now as bases are equal we can equate the powers,
 \[ \Rightarrow n - \dfrac{1}{2} = 0\]
Now we will add \[\dfrac{1}{2}\] to both sides we get,
 \[ \Rightarrow n - \dfrac{1}{2} + \dfrac{1}{2} = 0 + \dfrac{1}{2}\],
Now we will simplify, then we will get,
 \[ \Rightarrow n = \dfrac{1}{2}\],
 \[\therefore \] The value of \[n\] is \[\dfrac{1}{2}\].

The correct option is C.

Note:The Geometric Mean (GM) is the average value or mean which indicates the middle tendency of the set of numbers by finding out the product of their values or numbers.
For a set of \[n\] data or numbers or observations, a geometric mean is the \[{n^{th}}\] root of their product. Basically, we just multiply all the numbers together and take the \[{n^{th}}\] root of the multiplied numbers, where \[n\] = the total number of data values.