
The expression \[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\] \[[a \ne b \ne 0]\] is (where a and \[{\rm{b}}\] are unequal non-zero numbers)
A. A.M. between a and b if \[n = - 1\]
B. G.M. between a and b if \[n = - \dfrac{1}{2}\]
C. H.M. between a and b if \[n = 0\]
D. All are correct
Answer
161.4k+ views
Hint: In this case, we have been given the expression \[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\]here \[[a \ne b \ne 0]\] and the terms ‘a’ and ‘b’ are non-zero unequal numbers and we are asked to determine the value of ‘n’. For that, we have to equate the given expression to A.P formula to determine the value of ‘n’ and equate the same to the formula of H.M and also equate the expression to G.P \[\sqrt {ab} \] to get the value of n as a desired solution for the given problem.
Formula Used: A.M of two terms ‘a’ and ‘b’
\[\dfrac{{a + b}}{2}\]
G.M of two terms ‘a’ and ‘b’
\[\sqrt {ab} \]
H.M. of two terms ‘a’ and ‘b’
\[\dfrac{{2ab}}{{a + b}}\]
Complete step by step solution: A.M of two terms ‘a’ and ‘b’
\[\dfrac{{a + b}}{2}\]
G.M of two terms ‘a’ and ‘b’
\[\sqrt {ab} \]
H.M. of two terms ‘a’ and ‘b’
\[\dfrac{{2ab}}{{a + b}}\]
COMPLETE STEP-BY-STEP SOLUTION:
We have been provided in the question that,
\[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\]
Here, \[[a \ne b \ne 0]\]
And a and \[{\rm{b}}\] are unequal non-zero numbers
Now, let us consider that,
\[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \dfrac{{a + b}}{2}\]
Now, on applying cross product method to the above equation, we get
We have been already known that if bases are multiplied then the power should be added, we get \[ \Rightarrow 2{a^{n + 1}} + 2{b^{n + 1}} = {a^{n + 1}} + {b^{n + 1}} + a{b^n} + b{a^n}\]
Now, we have to cancel the similar terms on either sides of the equation, we get
\[{a^{n + 1}} - {a^n}b + {b^n} + 1 - a{b^n} = 0\]
Now, we have to reduce the above equation, we have
\[ \Rightarrow (a - b)({a^n} - {b^n}) = 0\]
We have been provided the condition that ‘a’ and ‘b’ are non-zero unequal numbers.
So,
\[ \Rightarrow {a^n} = {b^n}\]
Now, let us consider that
\[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \sqrt {ab} \]
Now, on applying cross product method to the above equation, we get
\[ \Rightarrow {a^{n + 1}} + {b^{n + 1}} = {a^{n + \dfrac{1}{2}}}{b^{\dfrac{1}{2}}} + {a^{\dfrac{1}{2}}}{b^{n + \dfrac{1}{2}}}\]
Now, we have to reduce the above equation using formula, we get
\[\left( {{a^{n + \dfrac{1}{2}}} - {b^{n + \dfrac{1}{2}}}} \right)(\sqrt a - \sqrt b ) = 0\]
Now, we have to move \[(\sqrt a - \sqrt b )\] to the other side of the equation, we get
\[ \Rightarrow {a^{n + \dfrac{1}{2}}} - {b^{n + \dfrac{1}{2}}} = 0\]
The above will be true if
\[n + \dfrac{1}{2} = 0\]
On solving for \[n\] we get
\[ \Rightarrow n = - \dfrac{1}{2}\]
Now, let us consider that,
\[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \dfrac{{2ab}}{{a + b}}\]
Now, on applying cross product method to the above equation, we get
\[ \Rightarrow {a^{n + 2}} + {a^{n + 1}}b + a{b^{n + 1}} + {b^{n + 2}} = 2{a^{n + 1}}b + 2a{b^{n + 1}}\]
Now, we have to reduce the above expression using formula we get
\[ \Rightarrow (a - b)\left( {{a^{n + 1}} - {b^{n + 1}}} \right) = 0\]
We have been already known that if bases are multiplied then the power should be added, we get \[ \Rightarrow {a^{n + 1}} - {b^{n + 1}} = 0\]
Now, we have obtained that,
\[ \Rightarrow n = - 1\]
Therefore, G.M. between a and b if \[n = - \dfrac{1}{2}\]
Option ‘B’ is correct
Note: Students are likely to make mistakes in these types of problems because it includes more formulas to be remembered. Applying wrong formula will lead to wrong solution. So one should keep in mind that A.M between two terms is \[\dfrac{{a + b}}{2}\] then H.M between two terms is \[\dfrac{{2ab}}{{a + b}}\] and G.M between two terms is \[\sqrt {ab} \].
Formula Used: A.M of two terms ‘a’ and ‘b’
\[\dfrac{{a + b}}{2}\]
G.M of two terms ‘a’ and ‘b’
\[\sqrt {ab} \]
H.M. of two terms ‘a’ and ‘b’
\[\dfrac{{2ab}}{{a + b}}\]
Complete step by step solution: A.M of two terms ‘a’ and ‘b’
\[\dfrac{{a + b}}{2}\]
G.M of two terms ‘a’ and ‘b’
\[\sqrt {ab} \]
H.M. of two terms ‘a’ and ‘b’
\[\dfrac{{2ab}}{{a + b}}\]
COMPLETE STEP-BY-STEP SOLUTION:
We have been provided in the question that,
\[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\]
Here, \[[a \ne b \ne 0]\]
And a and \[{\rm{b}}\] are unequal non-zero numbers
Now, let us consider that,
\[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \dfrac{{a + b}}{2}\]
Now, on applying cross product method to the above equation, we get
We have been already known that if bases are multiplied then the power should be added, we get \[ \Rightarrow 2{a^{n + 1}} + 2{b^{n + 1}} = {a^{n + 1}} + {b^{n + 1}} + a{b^n} + b{a^n}\]
Now, we have to cancel the similar terms on either sides of the equation, we get
\[{a^{n + 1}} - {a^n}b + {b^n} + 1 - a{b^n} = 0\]
Now, we have to reduce the above equation, we have
\[ \Rightarrow (a - b)({a^n} - {b^n}) = 0\]
We have been provided the condition that ‘a’ and ‘b’ are non-zero unequal numbers.
So,
\[ \Rightarrow {a^n} = {b^n}\]
Now, let us consider that
\[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \sqrt {ab} \]
Now, on applying cross product method to the above equation, we get
\[ \Rightarrow {a^{n + 1}} + {b^{n + 1}} = {a^{n + \dfrac{1}{2}}}{b^{\dfrac{1}{2}}} + {a^{\dfrac{1}{2}}}{b^{n + \dfrac{1}{2}}}\]
Now, we have to reduce the above equation using formula, we get
\[\left( {{a^{n + \dfrac{1}{2}}} - {b^{n + \dfrac{1}{2}}}} \right)(\sqrt a - \sqrt b ) = 0\]
Now, we have to move \[(\sqrt a - \sqrt b )\] to the other side of the equation, we get
\[ \Rightarrow {a^{n + \dfrac{1}{2}}} - {b^{n + \dfrac{1}{2}}} = 0\]
The above will be true if
\[n + \dfrac{1}{2} = 0\]
On solving for \[n\] we get
\[ \Rightarrow n = - \dfrac{1}{2}\]
Now, let us consider that,
\[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \dfrac{{2ab}}{{a + b}}\]
Now, on applying cross product method to the above equation, we get
\[ \Rightarrow {a^{n + 2}} + {a^{n + 1}}b + a{b^{n + 1}} + {b^{n + 2}} = 2{a^{n + 1}}b + 2a{b^{n + 1}}\]
Now, we have to reduce the above expression using formula we get
\[ \Rightarrow (a - b)\left( {{a^{n + 1}} - {b^{n + 1}}} \right) = 0\]
We have been already known that if bases are multiplied then the power should be added, we get \[ \Rightarrow {a^{n + 1}} - {b^{n + 1}} = 0\]
Now, we have obtained that,
\[ \Rightarrow n = - 1\]
Therefore, G.M. between a and b if \[n = - \dfrac{1}{2}\]
Option ‘B’ is correct
Note: Students are likely to make mistakes in these types of problems because it includes more formulas to be remembered. Applying wrong formula will lead to wrong solution. So one should keep in mind that A.M between two terms is \[\dfrac{{a + b}}{2}\] then H.M between two terms is \[\dfrac{{2ab}}{{a + b}}\] and G.M between two terms is \[\sqrt {ab} \].
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