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Question Number 12566 by JAZAR last updated on 25/Apr/17

$$wegive{U}_1,U_2,U_{3}thetermsofageometricsequence$$$$.Determine{U}_1,U_2,U_{3}suchthat:$$$$$$$$\{\begin{array}{l}{U}_1.U_2.U_3=64\\ U_1^2+U_2^2+U_3^2=84\end{array}$$$$$$

Answered by sandy_suhendra last updated on 25/Apr/17

$${\mathrm{U}}_1=\mathrm{a}$$$${\mathrm{U}}_2=\mathrm{ar}$$$${\mathrm{U}}_3={\mathrm{ar}}^2$$$${\mathrm{U}}_1.{\mathrm{U}}_2.{\mathrm{U}}_3=64$$$$\mathrm{a}.\mathrm{ar}.{\mathrm{ar}}^2=64$$$${\mathrm{a}}^3{\mathrm{r}}^3=64$$$$\mathrm{ar}=4={\mathrm{U}}_2\Rightarrow \mathrm{a}=\frac{4}{\mathrm{r}}={\mathrm{U}}_1$$$${\mathrm{U}}_3={\mathrm{ar}}^2=\mathrm{ar}.\mathrm{r}=4\mathrm{r}$$$$$$$${\mathrm{U}}_1^2+{\mathrm{U}}_2^2+{\mathrm{U}}_3^2=84$$$${\left(\frac{4}{\mathrm{r}}\right)}^2+4^2+{\left(4\mathrm{r}\right)}^2=84$$$$\frac{16}{{\mathrm{r}}^2}+16+{16\mathrm{r}}^2=84$$$${16\mathrm{r}}^2-68+\frac{16}{{\mathrm{r}}^2}=0\mathrm{multiplied}\mathrm{by}\left(\frac{{\mathrm{r}}^2}{4}\right)$$$${4\mathrm{r}}^4-{17\mathrm{r}}^2+4=0$$$$({4\mathrm{r}}^2-1)({\mathrm{r}}^2-4)=0$$$$\mathrm{r}=\pm \frac{1}{2}\mathrm{or}\mathrm{r}=\pm 2$$$$\mathrm{if}\mathrm{r}=-\frac{1}{2}\Rightarrow {\mathrm{U}}_1=-8;{\mathrm{U}}_2=4;{\mathrm{U}}_3=-2$$$$\mathrm{if}\mathrm{r}=\frac{1}{2}\Rightarrow {\mathrm{U}}_1=8;{\mathrm{U}}_2=4;{\mathrm{U}}_3=2$$$$\mathrm{if}\mathrm{r}=-2\Rightarrow {\mathrm{U}}_1=-2;{\mathrm{U}}_2=4;{\mathrm{U}}_3=-8$$$$\mathrm{if}\mathrm{r}=2\Rightarrow {\mathrm{U}}_1=2;{\mathrm{U}}_2=4;{\mathrm{U}}_3=8$$$$$$

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