
If \[\dfrac{{\left( {{a^{n + 1}} + {b^{n + 1}}} \right)}}{{\left( {{a^n} + {b^n}} \right)}}\] be the arithmetic mean of \[a\] and \[b\], then what is the value of \[n\]?
A. 1
B. \[ - 1\]
C. 0
D. None of these
Answer
232.8k+ views
Hint:First, calculate the arithmetic mean of \[a\] and \[b\]. Then equate the given value with the arithmetic mean. Solve the equation to reach the required answer.
Formula Used:
An arithmetic mean of the two numbers \[x\] and \[y\] is: \[\dfrac{{x + y}}{2}\]
\[a{}^m \times {a^n} = {a^{m + n}}\]
Complete step by step solution:
Given:
The arithmetic mean of \[a\] and \[b\] is \[\dfrac{{\left( {{a^{n + 1}} + {b^{n + 1}}} \right)}}{{\left( {{a^n} + {b^n}} \right)}}\].
Let’s calculate the arithmetic mean of \[a\] and \[b\].
\[A.M. = \dfrac{{a + b}}{2}\]
Now equate it with the given value.
We get,
\[\dfrac{{\left( {{a^{n + 1}} + {b^{n + 1}}} \right)}}{{\left( {{a^n} + {b^n}} \right)}} = \dfrac{{a + b}}{2}\]
Cross multiply.
\[2\left( {{a^{n + 1}} + {b^{n + 1}}} \right) = \left( {a + b} \right)\left( {{a^n} + {b^n}} \right)\]
Factor out the common terms.
\[2{a^{n + 1}} + 2{b^{n + 1}} = a \times {a^n} + a \times {b^n} + b \times {a^n} + b \times {b^n}\]
\[ \Rightarrow \]\[2{a^{n + 1}} + 2{b^{n + 1}} = {a^{n + 1}} + {b^{n + 1}} + a{b^n} + b{a^n}\]
Cancel out the common terms.
\[{a^{n + 1}} + {b^{n + 1}} = a{b^n} + b{a^n}\]
Now apply the exponent property \[a{}^m \times {a^n} = {a^{m + n}}\].
\[a{a^n} + b{b^n} = a{b^n} + b{a^n}\]
\[ \Rightarrow \]\[a{a^n} - b{a^n} = a{b^n} - b{b^n}\]
Simplify the above equation.
\[\left( {a - b} \right){a^n} = \left( {a - b} \right){b^n}\]
Cancel out the common terms from both sides.
\[{a^n} = {b^n}\]
\[ \Rightarrow \]\[\dfrac{{{a^n}}}{{{b^n}}} = 1\]
\[ \Rightarrow \]\[{\left( {\dfrac{a}{b}} \right)^n} = 1\]
\[ \Rightarrow \]\[{\left( {\dfrac{a}{b}} \right)^n} = {\left( {\dfrac{a}{b}} \right)^0}\] [ Zero exponent rule]
Since \[a\] and \[b\] are two different numbers.
So, this is possible only when \[n = 0\].
Hence the correct option is C.
Note: The arithmetic mean of numbers is the ratio of the sum of numbers to the total number of numbers. It is also called an average of the numbers.
Zero exponent rule: If the exponent of any number is 0, then the output is 1.
Formula Used:
An arithmetic mean of the two numbers \[x\] and \[y\] is: \[\dfrac{{x + y}}{2}\]
\[a{}^m \times {a^n} = {a^{m + n}}\]
Complete step by step solution:
Given:
The arithmetic mean of \[a\] and \[b\] is \[\dfrac{{\left( {{a^{n + 1}} + {b^{n + 1}}} \right)}}{{\left( {{a^n} + {b^n}} \right)}}\].
Let’s calculate the arithmetic mean of \[a\] and \[b\].
\[A.M. = \dfrac{{a + b}}{2}\]
Now equate it with the given value.
We get,
\[\dfrac{{\left( {{a^{n + 1}} + {b^{n + 1}}} \right)}}{{\left( {{a^n} + {b^n}} \right)}} = \dfrac{{a + b}}{2}\]
Cross multiply.
\[2\left( {{a^{n + 1}} + {b^{n + 1}}} \right) = \left( {a + b} \right)\left( {{a^n} + {b^n}} \right)\]
Factor out the common terms.
\[2{a^{n + 1}} + 2{b^{n + 1}} = a \times {a^n} + a \times {b^n} + b \times {a^n} + b \times {b^n}\]
\[ \Rightarrow \]\[2{a^{n + 1}} + 2{b^{n + 1}} = {a^{n + 1}} + {b^{n + 1}} + a{b^n} + b{a^n}\]
Cancel out the common terms.
\[{a^{n + 1}} + {b^{n + 1}} = a{b^n} + b{a^n}\]
Now apply the exponent property \[a{}^m \times {a^n} = {a^{m + n}}\].
\[a{a^n} + b{b^n} = a{b^n} + b{a^n}\]
\[ \Rightarrow \]\[a{a^n} - b{a^n} = a{b^n} - b{b^n}\]
Simplify the above equation.
\[\left( {a - b} \right){a^n} = \left( {a - b} \right){b^n}\]
Cancel out the common terms from both sides.
\[{a^n} = {b^n}\]
\[ \Rightarrow \]\[\dfrac{{{a^n}}}{{{b^n}}} = 1\]
\[ \Rightarrow \]\[{\left( {\dfrac{a}{b}} \right)^n} = 1\]
\[ \Rightarrow \]\[{\left( {\dfrac{a}{b}} \right)^n} = {\left( {\dfrac{a}{b}} \right)^0}\] [ Zero exponent rule]
Since \[a\] and \[b\] are two different numbers.
So, this is possible only when \[n = 0\].
Hence the correct option is C.
Note: The arithmetic mean of numbers is the ratio of the sum of numbers to the total number of numbers. It is also called an average of the numbers.
Zero exponent rule: If the exponent of any number is 0, then the output is 1.
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