If $$\dfrac{a^n+b^n}{a^{n-1}+b^{n-1}}$$is the A.M. between a and b, then find the value of n.

Asked by Abhisek | 1 year ago |  77

#### 1 Answer

##### Solution :-

The A.M between a and b is given by, $$\dfrac{(a + b)}{2}$$

Then according to the question,

$$\dfrac{a + b}{2} = \dfrac{a^n + b^n}{a^{n-1}+b^{n-1}}$$

2(an+bn)=(a+b)$$(a^{n−1}+b^{n−1)}$$

⇒2an+2b$$a (a^{n−1}+b^{n−1})+b (a^{n−1}+b^{n−1})$$

$$2a^n+2b^n = a^n+ab^{n−1}+ba^{n−1}+b^n$$

$$2a^n+2b^n−a^n−ab^{n−1}−ba^{n−1}−b^n=0$$

$$a^n−ba^{n−1}+b^n−ab^{n−1}=0$$

$$a^{n−1}(a−b)−b^{n−1}(a−b)=0$$

$$a^{n−1}−b^{n−1}=0\; or\; a−b=0$$

$$a^{n−1}=b^{n−1} \;or\; a=b\; but\; a\neq b$$

$$\dfrac{a^{n-1}}{b^{n-1}}=1$$

$$( \dfrac{a}{b})^{n-1}= ( \dfrac{a}{b})^0$$

Thus, the value of n is 1.

Answered by Pragya Singh | 1 year ago

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