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Question Number 28327 by Rasheed.Sindhi last updated on 24/Jan/18

$$\mathrm{Given}\mathrm{that}\frac{a^{n+1}+b^{n+1}}{a^n+b^n}\mathrm{is}\mathrm{AM}\mathrm{between}a$$$$\mathrm{and}b,\mathrm{where}a\ne b\wedge a,b\ne 0;\mathrm{find}\mathrm{out}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{n}.$$

Answered by mrW2 last updated on 24/Jan/18

$$\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\frac{a+b}{2}$$$$2(a^{n+1}+b^{n+1})=(a+b)(a^n+b^n)$$$$2a^{n+1}+2b^{n+1}=a^{n+1}+b^{n+1}+{ab}^n+{ba}^n$$$$a^{n+1}+b^{n+1}={ab}^n+{ba}^n$$$$a^n(a-b)=b^n(a-b)$$$$sincea\ne b,a-b\ne 0and\frac{a}{b}\ne 1$$$$\Rightarrow a^n=b^n$$$$\Rightarrow {\left(\frac{a}{b}\right)}^n=1$$$$\Rightarrow n=\frac{\log 1}{\log \left(\frac{a}{b}\right)}=\frac{0}{\log \left(\frac{a}{b}\right)}=0$$

Commented by Rasheed.Sindhi last updated on 25/Jan/18

$$\mathbb{T}\mathrm{h}\alpha \mathrm{n}\mathbb{k}\mathrm{s}\mathrm{a}\mathrm{lot}\mathrm{Sir}!$$

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