
If \[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\] be the A.M of a and b, then \[n = \]
a). \[1\]
b). \[ - 1\]
c). \[0\]
d). None of these
Answer
491.7k+ views
Hint: Here we are asked to find the value of the term \[n\]. They have given that the given expression is the arithmetic mean of two terms \[a\] and\[b\]. The arithmetic mean of two terms can be calculated by dividing the sum of those two terms by two, equating this and the given A.M and simplifying it will yield the value \[n\].
Formula: Formulas that we need to know:
Arithmetic mean of \[a\] and \[b\]\[ = \dfrac{{a + b}}{2}\]
\[{a^x}.{a^y} = {a^{x + y}}\]
Complete step-by-step solution:
It is given that \[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\] is the arithmetic mean of the terms \[a\] and \[b\]. We aim to find the value of\[n\]
We know that the arithmetic mean of two terms (say \[a\]and\[b\]) is nothing but the average of those two terms, that is \[\dfrac{{a + b}}{2}\].
Thus, from the given data we get\[\dfrac{{a + b}}{2} = \dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\]. Now let us simplify this expression to find the value of the term\[n\].
Consider the expression, \[\dfrac{{a + b}}{2} = \dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\]
On cross multiplying the above expression, we get
\[ \Rightarrow \left( {{a^n} + {b^n}} \right)\left( {a + b} \right) = 2\left( {{a^{n + 1}} + {b^{n + 1}}} \right)\]
Let’s multiply the two terms on the left-hand side.
\[ \Rightarrow {a^n}a + {a^n}b + {b^n}a + {b^n}b = 2\left( {{a^{n + 1}} + {b^{n + 1}}} \right)\]
Using the formula \[{a^x}.{a^y} = {a^{x + y}}\] we get
\[ \Rightarrow {a^{n + 1}} + {a^n}b + {b^n}a + {b^{n + 1}} = 2\left( {{a^{n + 1}} + {b^{n + 1}}} \right)\]
Now let us multiply two on the right-hand side.
\[ \Rightarrow {a^{n + 1}} + {a^n}b + {b^n}a + {b^{n + 1}} = 2{a^{n + 1}} + 2{b^{n + 1}}\]
Let’s shift the terms \[{a^{n + 1}}\] and \[{b^{n + 1}}\] to the other side.
\[ \Rightarrow {a^n}b + {b^n}a = {a^{n + 1}} + {b^{n + 1}}\]
On simplifying the above, we get
\[ \Rightarrow {a^n}b + {b^n}a = {a^{n + 1}} + {b^{n + 1}}\]
Now let’s shift the terms \[{b^n}a\]and \[{a^{n + 1}}\] to the other side.
\[ \Rightarrow {a^n}b - {a^{n + 1}} = {b^{n + 1}} - {b^n}a\]
Let’s split the terms using the formula \[{a^x}.{a^y} = {a^{x + y}}\]
\[ \Rightarrow {a^n}b - {a^n}.a = {b^n}.b - {b^n}a\]
Now let’s take \[{a^n}\] and \[{b^n}\] commonly out from the left and right side respectively.
\[ \Rightarrow {a^n}\left( {b - a} \right) = {b^n}\left( {b - a} \right)\]
On simplifying this we get
\[ \Rightarrow {a^n} = {b^n}\]
\[ \Rightarrow \dfrac{{{a^n}}}{{{b^n}}} = 1\]
\[n = 0\]
\[ \Rightarrow n = 0\]
Thus, we have found the value\[n\]
Now let us see the options to find the right answer.
Option (a) \[1\] is an incorrect answer as we got \[n = 0\] in our calculation.
Option (b) \[ - 1\] is an incorrect answer as we got \[n = 0\] in our calculation.
Option (c) \[0\] is the correct answer as we got the same value in our calculation above.
Option (d) None of the above is the incorrect answer as we got the option (c) as the correct answer.
Hence, option (c) \[0\] is the correct answer.
Note: The arithmetic mean of any number of data can be calculated as the sum of the given data divided by the total number of the data. In the above calculation, we got that \[{\left( {\dfrac{a}{b}} \right)^n} = 1 \Rightarrow n = 0\] this is because anything raised to the power \[0\] equals \[1\] thus we got\[n = 0\].
Formula: Formulas that we need to know:
Arithmetic mean of \[a\] and \[b\]\[ = \dfrac{{a + b}}{2}\]
\[{a^x}.{a^y} = {a^{x + y}}\]
Complete step-by-step solution:
It is given that \[\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\] is the arithmetic mean of the terms \[a\] and \[b\]. We aim to find the value of\[n\]
We know that the arithmetic mean of two terms (say \[a\]and\[b\]) is nothing but the average of those two terms, that is \[\dfrac{{a + b}}{2}\].
Thus, from the given data we get\[\dfrac{{a + b}}{2} = \dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\]. Now let us simplify this expression to find the value of the term\[n\].
Consider the expression, \[\dfrac{{a + b}}{2} = \dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}\]
On cross multiplying the above expression, we get
\[ \Rightarrow \left( {{a^n} + {b^n}} \right)\left( {a + b} \right) = 2\left( {{a^{n + 1}} + {b^{n + 1}}} \right)\]
Let’s multiply the two terms on the left-hand side.
\[ \Rightarrow {a^n}a + {a^n}b + {b^n}a + {b^n}b = 2\left( {{a^{n + 1}} + {b^{n + 1}}} \right)\]
Using the formula \[{a^x}.{a^y} = {a^{x + y}}\] we get
\[ \Rightarrow {a^{n + 1}} + {a^n}b + {b^n}a + {b^{n + 1}} = 2\left( {{a^{n + 1}} + {b^{n + 1}}} \right)\]
Now let us multiply two on the right-hand side.
\[ \Rightarrow {a^{n + 1}} + {a^n}b + {b^n}a + {b^{n + 1}} = 2{a^{n + 1}} + 2{b^{n + 1}}\]
Let’s shift the terms \[{a^{n + 1}}\] and \[{b^{n + 1}}\] to the other side.
\[ \Rightarrow {a^n}b + {b^n}a = {a^{n + 1}} + {b^{n + 1}}\]
On simplifying the above, we get
\[ \Rightarrow {a^n}b + {b^n}a = {a^{n + 1}} + {b^{n + 1}}\]
Now let’s shift the terms \[{b^n}a\]and \[{a^{n + 1}}\] to the other side.
\[ \Rightarrow {a^n}b - {a^{n + 1}} = {b^{n + 1}} - {b^n}a\]
Let’s split the terms using the formula \[{a^x}.{a^y} = {a^{x + y}}\]
\[ \Rightarrow {a^n}b - {a^n}.a = {b^n}.b - {b^n}a\]
Now let’s take \[{a^n}\] and \[{b^n}\] commonly out from the left and right side respectively.
\[ \Rightarrow {a^n}\left( {b - a} \right) = {b^n}\left( {b - a} \right)\]
On simplifying this we get
\[ \Rightarrow {a^n} = {b^n}\]
\[ \Rightarrow \dfrac{{{a^n}}}{{{b^n}}} = 1\]
\[n = 0\]
\[ \Rightarrow n = 0\]
Thus, we have found the value\[n\]
Now let us see the options to find the right answer.
Option (a) \[1\] is an incorrect answer as we got \[n = 0\] in our calculation.
Option (b) \[ - 1\] is an incorrect answer as we got \[n = 0\] in our calculation.
Option (c) \[0\] is the correct answer as we got the same value in our calculation above.
Option (d) None of the above is the incorrect answer as we got the option (c) as the correct answer.
Hence, option (c) \[0\] is the correct answer.
Note: The arithmetic mean of any number of data can be calculated as the sum of the given data divided by the total number of the data. In the above calculation, we got that \[{\left( {\dfrac{a}{b}} \right)^n} = 1 \Rightarrow n = 0\] this is because anything raised to the power \[0\] equals \[1\] thus we got\[n = 0\].
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