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Question Number 147328 by Mrsof last updated on 19/Jul/21

Commented by Mrsof last updated on 19/Jul/21

$$whatstherightanswer$$

Answered by Olaf_Thorendsen last updated on 19/Jul/21

$$x_{n+1}=\frac{1}{2}{x}_n+2,x_1=8\left(1\right)$$$$$$$$\mathrm{Let}{y}_n=x_n-4\Rightarrow y_1=x_1-4=4$$$$\mathrm{and}\left(1\right):y_{n+1}+4=\frac{1}{2}(x_n+4)+2$$$$y_{n+1}=\frac{1}{2}{y}_n$$$$y_n=y_1{\left(\frac{1}{2}\right)}^{n-1}=4{\left(\frac{1}{2}\right)}^{n-1}={\left(\frac{1}{2}\right)}^{n-3}$$$$\Rightarrow x_n=y_n+4={\left(\frac{1}{2}\right)}^{n-3}+4$$$$\underset{n\to \mathrm{\infty }}{\lim }{x}_n=4$$

Commented by Mrsof last updated on 20/Jul/21

$$thankyousircanyougivemethenameof$$$$thisproplems$$

Commented by Olaf_Thorendsen last updated on 20/Jul/21

$$\mathrm{arithmetico}\mathrm{geometric}\mathrm{sequence}$$

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