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Question

For what value of $$n$$ so that $$\dfrac { { a }^{ n+1 }+{ b }^{ n+1 } }{ { a }^{ n }+{ b }^{ n } } $$ is the geometric mean of $$a$$ and $$b$$


Solution

We know that the geometric mean between $$a$$ and $$b$$ is $$=\sqrt{ab}$$
G.M between $$a$$ and $$b=\dfrac{{a}^{n+1}+{b}^{n+1}}{{a}^{n}+{b}^{n}}$$
$$\sqrt{ab}=\dfrac{{a}^{n+1}+{b}^{n+1}}{{a}^{n}+{b}^{n}}$$
$$\Rightarrow {a}^{\frac{1}{2}}{b}^{\frac{1}{2}}\left({a}^{n}+{b}^{n}\right)={a}^{n+1}+{b}^{n+1}$$
$$\Rightarrow {a}^{n+\frac{1}{2}}{b}^{\frac{1}{2}}+{a}^{\frac{1}{2}}{b}^{n+\frac{1}{2}}={a}^{n+1}+{b}^{n+1}$$
$$\Rightarrow {a}^{n+\frac{1}{2}}{b}^{\frac{1}{2}}-{a}^{n+1}={b}^{n+1}-{a}^{\frac{1}{2}}{b}^{n+\frac{1}{2}}$$
$$\Rightarrow {a}^{n+\frac{1}{2}}\left({b}^{\frac{1}{2}}-{a}^{\frac{1}{2}}\right)={b}^{n+\frac{1}{2}}\left({b}^{\frac{1}{2}}-{a}^{\frac{1}{2}}\right)$$ since $$\dfrac{1}{2}+\dfrac{1}{2}=1$$
$$\Rightarrow {a}^{n+\frac{1}{2}}={b}^{n+\frac{1}{2}}$$
$$\Rightarrow {\left(\dfrac{a}{b}\right)}^{n+\frac{1}{2}}=1$$
$$\Rightarrow {\left(\dfrac{a}{b}\right)}^{n+\frac{1}{2}}={\left(\dfrac{a}{b}\right)}^{0}$$
Comparing the powers, we get
$$n+\dfrac{1}{2}=0$$
$$\therefore n=\dfrac{-1}{2}$$

Mathematics

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