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Question Number 7525 by Little last updated on 02/Sep/16

The sum of the squares of three distinct  real numbers which are in GP is S^2 . If  their sum is α S, then

$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{squares}\:\mathrm{of}\:\mathrm{three}\:\mathrm{distinct} \\ $$$$\mathrm{real}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP}\:\mathrm{is}\:{S}^{\mathrm{2}} .\:\mathrm{If} \\ $$$$\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\alpha\:{S},\:\mathrm{then} \\ $$

Commented by Rasheed Soomro last updated on 04/Sep/16

Assumed that α is required.  Let a is the initial real number and r is  common ratio.  So three distinct real numbers are        a,ar,ar^2   Sum of squares: a^2 +a^2 r^2 +a^2 r^4 =S^( 2)     ...........(i)  Sum of real numbers: a+ar+ar^2 =αS  Or S=((a(1+r+r^2 ))/α)     ....................................(ii)  From (i)&(ii)      a^2 (1+r^2 +r^4 )=(((a(1+r+r^2 ))/α))^2             1+r^2 +r^4 =(((1+r+r^2 )^2 )/α^2 )              α^2 =(((1+r+r^2 )^2 )/(1+r^2 +r^4 ))

$${Assumed}\:{that}\:\alpha\:{is}\:{required}. \\ $$$${Let}\:{a}\:{is}\:{the}\:{initial}\:{real}\:{number}\:{and}\:{r}\:{is} \\ $$$${common}\:{ratio}. \\ $$$${So}\:{three}\:{distinct}\:{real}\:{numbers}\:{are} \\ $$$$\:\:\:\:\:\:{a},{ar},{ar}^{\mathrm{2}} \\ $$$${Sum}\:{of}\:{squares}:\:{a}^{\mathrm{2}} +{a}^{\mathrm{2}} {r}^{\mathrm{2}} +{a}^{\mathrm{2}} {r}^{\mathrm{4}} ={S}^{\:\mathrm{2}} \:\:\:\:...........\left({i}\right) \\ $$$${Sum}\:{of}\:{real}\:{numbers}:\:{a}+{ar}+{ar}^{\mathrm{2}} =\alpha{S} \\ $$$${Or}\:{S}=\frac{{a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)}{\alpha}\:\:\:\:\:....................................\left({ii}\right) \\ $$$${From}\:\left({i}\right)\&\left({ii}\right) \\ $$$$\:\:\:\:{a}^{\mathrm{2}} \left(\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} \right)=\left(\frac{{a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)}{\alpha}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} =\frac{\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\alpha^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\alpha^{\mathrm{2}} =\frac{\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} } \\ $$

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