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Question Number 64519 by necx1 last updated on 18/Jul/19
a,b,c is a geometric progression such  that  a+b+c=26  a^2 +b^2 +c^2 =364    find a,b,c
$${a},{b},{c}\:{is}\:{a}\:{geometric}\:{progression}\:{such} \\ $$$${that} \\ $$$${a}+{b}+{c}=\mathrm{26} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{364} \\ $$$$ \\ $$$${find}\:{a},{b},{c} \\ $$
Answered by MJS last updated on 19/Jul/19
a+ar+ar^2 =26 ⇒ a=((26)/(1+r+r^2 ))  a^2 +a^2 r^2 +a^2 r^4 =364 ⇒ a^2 =((364)/(1+r^2 +r^4 ))  ((26^2 )/((1+r+r^2 )^2 ))=((364)/(1+r^2 +r^4 ))  13(1+r^2 +r^4 )=7(1+r+r^2 )^2   6r^4 −14r^3 −8r^2 −14r+6=0  6(r−3)(r−(1/3))(x^2 +x+1)=0  ⇒ r=3∨r=(1/3)  r=3 ⇒ a=2  b=6  c=18  r=(1/3) ⇒ a=18  b=6  c=2
$${a}+{ar}+{ar}^{\mathrm{2}} =\mathrm{26}\:\Rightarrow\:{a}=\frac{\mathrm{26}}{\mathrm{1}+{r}+{r}^{\mathrm{2}} } \\ $$$${a}^{\mathrm{2}} +{a}^{\mathrm{2}} {r}^{\mathrm{2}} +{a}^{\mathrm{2}} {r}^{\mathrm{4}} =\mathrm{364}\:\Rightarrow\:{a}^{\mathrm{2}} =\frac{\mathrm{364}}{\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} } \\ $$$$\frac{\mathrm{26}^{\mathrm{2}} }{\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{364}}{\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} } \\ $$$$\mathrm{13}\left(\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} \right)=\mathrm{7}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{6}{r}^{\mathrm{4}} −\mathrm{14}{r}^{\mathrm{3}} −\mathrm{8}{r}^{\mathrm{2}} −\mathrm{14}{r}+\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{6}\left({r}−\mathrm{3}\right)\left({r}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{r}=\mathrm{3}\vee{r}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${r}=\mathrm{3}\:\Rightarrow\:{a}=\mathrm{2}\:\:{b}=\mathrm{6}\:\:{c}=\mathrm{18} \\ $$$${r}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{a}=\mathrm{18}\:\:{b}=\mathrm{6}\:\:{c}=\mathrm{2} \\ $$

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