Question Number 7525 by Little last updated on 02/Sep/16
$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{squares}\:\mathrm{of}\:\mathrm{three}\:\mathrm{distinct} \\ $$$$\mathrm{real}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP}\:\mathrm{is}\:{S}^{\mathrm{2}} .\:\mathrm{If} \\ $$$$\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\alpha\:{S},\:\mathrm{then} \\ $$
Commented by Rasheed Soomro last updated on 04/Sep/16
$${Assumed}\:{that}\:\alpha\:{is}\:{required}. \\ $$$${Let}\:{a}\:{is}\:{the}\:{initial}\:{real}\:{number}\:{and}\:{r}\:{is} \\ $$$${common}\:{ratio}. \\ $$$${So}\:{three}\:{distinct}\:{real}\:{numbers}\:{are} \\ $$$$\:\:\:\:\:\:{a},{ar},{ar}^{\mathrm{2}} \\ $$$${Sum}\:{of}\:{squares}:\:{a}^{\mathrm{2}} +{a}^{\mathrm{2}} {r}^{\mathrm{2}} +{a}^{\mathrm{2}} {r}^{\mathrm{4}} ={S}^{\:\mathrm{2}} \:\:\:\:………..\left({i}\right) \\ $$$${Sum}\:{of}\:{real}\:{numbers}:\:{a}+{ar}+{ar}^{\mathrm{2}} =\alpha{S} \\ $$$${Or}\:{S}=\frac{{a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)}{\alpha}\:\:\:\:\:………………………………\left({ii}\right) \\ $$$${From}\:\left({i}\right)\&\left({ii}\right) \\ $$$$\:\:\:\:{a}^{\mathrm{2}} \left(\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} \right)=\left(\frac{{a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)}{\alpha}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} =\frac{\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\alpha^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\alpha^{\mathrm{2}} =\frac{\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} } \\ $$