Question Number 147328 by Mrsof last updated on 19/Jul/21
Commented by Mrsof last updated on 19/Jul/21
$${whats}\:{the}\:{right}\:{answer} \\ $$
Answered by Olaf_Thorendsen last updated on 19/Jul/21
$${x}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}_{{n}} +\mathrm{2},\:{x}_{\mathrm{1}} \:=\:\mathrm{8}\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{Let}\:{y}_{{n}} \:=\:{x}_{{n}} −\mathrm{4}\:\Rightarrow\:{y}_{\mathrm{1}} \:=\:{x}_{\mathrm{1}} −\mathrm{4}\:=\:\mathrm{4} \\ $$$$\mathrm{and}\:\left(\mathrm{1}\right):\:\:{y}_{{n}+\mathrm{1}} +\mathrm{4}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}_{{n}} +\mathrm{4}\right)+\mathrm{2} \\ $$$${y}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{y}_{{n}} \\ $$$${y}_{{n}} \:=\:{y}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} =\:\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} =\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}−\mathrm{3}} \\ $$$$\Rightarrow\:{x}_{{n}} \:=\:{y}_{{n}} +\mathrm{4}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}−\mathrm{3}} +\mathrm{4} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{x}_{{n}} \:=\:\mathrm{4} \\ $$
Commented by Mrsof last updated on 20/Jul/21
$${thank}\:{you}\:{sir}\:{can}\:{you}\:{give}\:{me}\:{the}\:{name}\:{of} \\ $$$${this}\:{proplems} \\ $$
Commented by Olaf_Thorendsen last updated on 20/Jul/21
$$\mathrm{arithmetico}\:\mathrm{geometric}\:\mathrm{sequence} \\ $$