Given-a-b-and-c-3-real-numbers-which-satisfy-the-equation-a-b-c-312-c-a-192-Find-these-real-numbers-such-that-they-form-3-consecutive-terms-of-a-Geometric-Progression-

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Question Number 94573 by Ar Brandon last updated on 19/May/20

$$\mathrm{Given}\mathrm{a},\mathrm{b},\mathrm{and}\mathrm{c},3\mathrm{real}\mathrm{numbers}\mathrm{which}\mathrm{satisfy}$$$$\mathrm{the}\mathrm{equation}\{\begin{array}{l}\mathrm{a}+\mathrm{b}+\mathrm{c}=312\\ \mathrm{c}+\mathrm{a}=192\end{array}$$$$\mathrm{Find}\mathrm{these}\mathrm{real}\mathrm{numbers}\mathrm{such}\mathrm{that}\mathrm{they}\mathrm{form}$$$$3\mathrm{consecutive}\mathrm{terms}\mathrm{of}\mathrm{a}\mathrm{Geometric}\mathrm{Progression}.$$

Commented by mr W last updated on 19/May/20

$$a=\frac{b}{r}$$$$c=br$$$$ac=b^2=192$$$$\Rightarrow b=\pm 8\sqrt{3}$$$$\frac{b}{r}+b+br=312$$$$\frac{1}{r}+r+1=\frac{312}{b}$$$$r^2-(1-\frac{312}{b})r+1=0$$$$r=\frac{1}{2}[1-\frac{312}{b}\pm \sqrt{{(1-\frac{312}{b})}^2-4}]$$$$r=\frac{1}{2}[1\pm 13\sqrt{3}\pm \sqrt{{(1\pm 13\sqrt{3})}^2-4}]$$

Commented by Ar Brandon last updated on 19/May/20

Thanks Mr W

Answered by Rasheed.Sindhi last updated on 19/May/20

$$\mathrm{Given}\mathrm{a},\mathrm{b},\mathrm{and}\mathrm{c},3\mathrm{real}\mathrm{numbers}\mathrm{which}\mathrm{satisfy}$$$$\mathrm{the}\mathrm{equation}\{\begin{array}{l}\mathrm{a}+\mathrm{b}+\mathrm{c}=312\\ \mathrm{c}+\mathrm{a}=192\end{array}$$$$\mathrm{Find}\mathrm{these}\mathrm{real}\mathrm{numbers}\mathrm{such}\mathrm{that}\mathrm{they}\mathrm{form}$$$$3\mathrm{consecutive}\mathrm{terms}\mathrm{of}\mathrm{a}\mathrm{Geometric}\mathrm{Progression}.$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=312\wedge \mathrm{c}+\mathrm{a}=192$$$$\mathrm{b}=312-192=120$$$$\frac{120}{\mathrm{r}},120,120\mathrm{r}areinGP$$$$\frac{120}{\mathrm{r}}+120+120\mathrm{r}=312$$$$\frac{1}{\mathrm{r}}+1+\mathrm{r}=\frac{312}{120}=\frac{13}{5}$$$$5+5\mathrm{r}+{5\mathrm{r}}^2=13\mathrm{r}$$$${5\mathrm{r}}^2-8\mathrm{r}+5=0$$$$r=\frac{8\pm \sqrt{64-100}}{10}$$$$riscomplex$$$$\therefore a,carenotreal.$$$$$$

Commented by mr W last updated on 19/May/20

$$maybehemeanta\times c=192.$$

Commented by Ar Brandon last updated on 19/May/20

Thank you I arrived at the same situation.

Commented by Ar Brandon last updated on 19/May/20

Oh no, it's a+b. I guess there was a problem with the question.

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