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Worked Examples · Example 15

Q.If an+bnan−1+bn−1\dfrac{a^n+b^n}{a^{n-1}+b^{n-1}} is the arithmetic mean between aa and bb, then find the value of nn.

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Equating the given expression to the arithmetic mean a+b2\frac{a+b}{2} and simplifying isolates an−1=bn−1a^{n-1}=b^{n-1}, which forces n=1n=1.

The arithmetic mean of two numbers aa and bb is:

A.M.=a+b2\text{A.M.} = \frac{a+b}{2}

We are given that an+bnan−1+bn−1\dfrac{a^n+b^n}{a^{n-1}+b^{n-1}} equals this A.M., and must find nn.

  1. Set up the equation: an+bnan−1+bn−1=a+b2\dfrac{a^n+b^n}{a^{n-1}+b^{n-1}} = \dfrac{a+b}{2}.
  2. Cross-multiply: 2(an+bn)=(a+b)(an−1+bn−1)2(a^n+b^n) = (a+b)(a^{n-1}+b^{n-1}).
  3. Expand the right side: (a+b)(an−1+bn−1)=an+abn−1+an−1b+bn(a+b)(a^{n-1}+b^{n-1}) = a^n + ab^{n-1} + a^{n-1}b + b^n.
  4. So the equation becomes: 2an+2bn=an+bn+abn−1+an−1b2a^n+2b^n = a^n+b^n+ab^{n-1}+a^{n-1}b.
  5. Subtract an+bna^n+b^n from both sides: an+bn=abn−1+an−1ba^n+b^n = ab^{n-1}+a^{n-1}b.
  6. Rearrange: an−an−1b=abn−1−bna^n - a^{n-1}b = ab^{n-1} - b^n, i.e. an−1(a−b)=bn−1(a−b)a^{n-1}(a-b) = b^{n-1}(a-b). …

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