Question Number 160058 by Rasheed.Sindhi last updated on 24/Nov/21
$${Find}\:{n}\:{so}\:{that}\:\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }\:{may}\:{be} \\ $$$${the}\:{arithmetic}\:{mean}\:{between}\:{a} \\ $$$${and}\:{b}. \\ $$
Commented by Tinku Tara last updated on 24/Nov/21
$${n}=\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 24/Nov/21
$${Right}\:{sir},\:{process}\:{please}. \\ $$
Answered by Kunal12588 last updated on 24/Nov/21
$$\frac{{a}+{b}}{\mathrm{2}}=\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} } \\ $$$$\Rightarrow{a}^{{n}+\mathrm{1}} +{ab}^{{n}} +{a}^{{n}} {b}+{b}^{{n}+\mathrm{1}} =\mathrm{2}{a}^{{n}+\mathrm{1}} +\mathrm{2}{b}^{{n}+\mathrm{1}} \\ $$$$\Rightarrow{a}^{{n}} {a}−{ab}^{{n}} −{a}^{{n}} {b}+{b}^{{n}} {b}=\mathrm{0} \\ $$$$\Rightarrow{a}\left({a}^{{n}} −{b}^{{n}} \right)−{b}\left({a}^{{n}} −{b}^{{n}} \right)=\mathrm{0} \\ $$$$\Rightarrow\left({a}−{b}\right)\left({a}^{{n}} −{b}^{{n}} \right)=\mathrm{0} \\ $$$$\Rightarrow{a}={b}\:{or}\:{a}^{{n}} ={b}^{{n}} \\ $$$${a}^{{n}} ={b}^{{n}} \Rightarrow\left(\frac{{a}}{{b}}\right)^{{n}} =\mathrm{1}\Rightarrow{n}=\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 24/Nov/21
$$\mathcal{T}{han}\mathcal{X}\:{sir}! \\ $$