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Question: If the arithmetic mean of \(a\) and \(b\) is \(\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}\...

If the arithmetic mean of aa and bb is an+bnan−1+bn−1\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}, then find the value of nn.
A) 11
B) 00
C) −1 - 1
D) 22

Explanation

Solution

We will be going to use the general formula of the arithmetic mean and then simplifying the solutions by using the basic rules of indices, BODMAS or taking transpose of the desired equations. Algebraic and simplification rules will be used in simplifying the expression.

Complete step by step answer:
Here, we have given the expression as an+bnan−1+bn−1\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}where there is condition that ‘a’ and ‘b’ are the arithmetic mean of it.
Therefore, it can written as
an+bnan−1+bn−1=a+b2\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}} = \dfrac{{a + b}}{2}
Because, the parameters ‘a’ and ‘b’ equally dependent due to arithmetic condition.
Hence, simplifying the above equation by applying the basic rules learnt in earlier classes, we get
2(an+bn)=(a+b)(an−1+bn−1)2\left( {{a^n} + {b^n}} \right) = \left( {a + b} \right)\left( {{a^{n - 1}} + {b^{n - 1}}} \right)
2an+2bn=aan−1+abn−1+ban−1+bbn−12{a^n} + 2{b^n} = a{a^{n - 1}} + a{b^{n - 1}} + b{a^{n - 1}} + b{b^{n - 1}}
Since, deriving the equation as per the BODMAS condition which will give the exact and appropriate solution,
aan−1+abn−1−2an−2bn+ban−1+bbn−1=0a{a^{n - 1}} + a{b^{n - 1}} - 2{a^n} - 2{b^n} + b{a^{n - 1}} + b{b^{n - 1}} = 0
an+abn−1−2an−2bn+ban−1+bn=0{a^n} + a{b^{n - 1}} - 2{a^n} - 2{b^n} + b{a^{n - 1}} + {b^n} = 0
Here, we have used the rules of indices for multiplication, addition, etc…
abn−1−an−bn+ban−1=0a{b^{n - 1}} - {a^n} - {b^n} + b{a^{n - 1}} = 0
Since, to make the equation easy to solve the elimination method is the best way.
Hence, rearranging the terms, we get
abn−1−bn+ban−1−an=0a{b^{n - 1}} - {b^n} + b{a^{n - 1}} - {a^n} = 0
So, shifting the two values on R.H.S., we get
abn−1−bn=an−ban−1a{b^{n - 1}} - {b^n} = {a^n} - b{a^{n - 1}}
bn−1(a−b)=an−1(a−b){b^{n - 1}}\left( {a - b} \right) = {a^{n - 1}}\left( {a - b} \right)
Dividing the common terms, equation becomes
bn−1=an−1{b^{n - 1}} = {a^{n - 1}}
Again taking the R.H.S. term in L.H.S. side, we get
bn−1an−1=1\dfrac{{{b^{n - 1}}}}{{{a^{n - 1}}}} = 1
(Simplifying as per the convenience of the problem)
(ba)n−1=1{\left( {\dfrac{b}{a}} \right)^{n - 1}} = 1
… (Applying rules of indices)
Now,
Hence, to achieve the R.H.S. as 11, L.H.S. should be also 11
So,
As per rules of indices anything raised to zero is 11,
Therefore, n−1n - 1 must equal zero.
n−1=0 ∴n=1  n - 1 = 0 \\\ \therefore n = 1 \\\
Therefore, the required value of nn is 11. Option (A) is correct.
Note:
Considering the general nth term itself in the formula i.e. general term of the formula a+b2\dfrac{{a + b}}{2} should be simplified simply to get the desired value for the given expression provided. Use the BODMAS rule to solve the calculations. Care should be taken while doing the calculations as it changes the final answer.