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

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.

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

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

a=(b/r) c=br ac=b^2 =192 ⇒b=±8(√3) (b/r)+b+br=312 (1/r)+r+1=((312)/b) r^2 −(1−((312)/b))r+1=0 r=(1/2)[1−((312)/b)±(√((1−((312)/b))^2 −4))] r=(1/2)[1±13(√3)±(√((1±13(√3))^2 −4))]

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

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

Thanks Mr W

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

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. a+b+c=312 ∧ c+a=192 b=312−192=120 ((120)/r) ,120 ,120r are in GP ((120)/r) +120 +120r=312 (1/r) +1+r=((312)/(120))=((13)/5) 5+5r+5r^2 =13r 5r^2 −8r+5=0 r=((8±(√(64−100)))/(10)) r is complex ∴a,c are not real.

$$\mathrm{Given}\:\mathrm{a},\:\mathrm{b},\:\mathrm{and}\:\mathrm{c},\:\mathrm{3}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{satisfy} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\begin{cases}{\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{312}}\\{\mathrm{c}+\mathrm{a}=\mathrm{192}}\end{cases} \\ $$$$\mathrm{Find}\:\mathrm{these}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\mathrm{they}\:\mathrm{form} \\ $$$$\mathrm{3}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{Geometric}\:\mathrm{Progression}. \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{312}\:\:\wedge\:\mathrm{c}+\mathrm{a}=\mathrm{192} \\ $$$$\mathrm{b}=\mathrm{312}−\mathrm{192}=\mathrm{120} \\ $$$$\frac{\mathrm{120}}{\mathrm{r}}\:,\mathrm{120}\:,\mathrm{120r}\:{are}\:{in}\:{GP} \\ $$$$\frac{\mathrm{120}}{\mathrm{r}}\:+\mathrm{120}\:+\mathrm{120r}=\mathrm{312} \\ $$$$\frac{\mathrm{1}}{\mathrm{r}}\:+\mathrm{1}+\mathrm{r}=\frac{\mathrm{312}}{\mathrm{120}}=\frac{\mathrm{13}}{\mathrm{5}} \\ $$$$\:\:\mathrm{5}+\mathrm{5r}+\mathrm{5r}^{\mathrm{2}} =\mathrm{13r} \\ $$$$\:\:\mathrm{5r}^{\mathrm{2}} −\mathrm{8r}+\mathrm{5}=\mathrm{0} \\ $$$$\:\:\:{r}=\frac{\mathrm{8}\pm\sqrt{\mathrm{64}−\mathrm{100}}}{\mathrm{10}} \\ $$$${r}\:{is}\:{complex} \\ $$$$\:\therefore{a},{c}\:{are}\:{not}\:{real}. \\ $$$$ \\ $$

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

maybe he meant a×c=192.

$${maybe}\:{he}\:{meant}\:{a}×{c}=\mathrm{192}. \\ $$

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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