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Question Number 9484 by Joel575 last updated on 10/Dec/16

S(r) = 12 + 12r^1  + 12r^2  + 12r^3  + 12r^4  + ...  with −1 < r < 1    If  S(a) . S(−a) = 2016, with −1 < a < 1  What is the value of  S(a) + S(−a) ?

$$\mathrm{S}\left({r}\right)\:=\:\mathrm{12}\:+\:\mathrm{12}{r}^{\mathrm{1}} \:+\:\mathrm{12}{r}^{\mathrm{2}} \:+\:\mathrm{12}{r}^{\mathrm{3}} \:+\:\mathrm{12}{r}^{\mathrm{4}} \:+\:... \\ $$ $$\mathrm{with}\:−\mathrm{1}\:<\:{r}\:<\:\mathrm{1} \\ $$ $$ \\ $$ $$\mathrm{If}\:\:\mathrm{S}\left({a}\right)\:.\:\mathrm{S}\left(−{a}\right)\:=\:\mathrm{2016},\:\mathrm{with}\:−\mathrm{1}\:<\:{a}\:<\:\mathrm{1} \\ $$ $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{S}\left({a}\right)\:+\:\mathrm{S}\left(−{a}\right)\:? \\ $$

Answered by mrW last updated on 10/Dec/16

S(r)=((12)/(1−r))  S(a)=((12)/(1−a))  S(−a)=((12)/(1+a))  S(a)∙S(−a)=((12)/(1−a))×((12)/(1+a))=2016  ((144)/(1−a^2 ))=2016  (1/(1−a^2 ))=((2016)/(144))=14  a=±(√((13)/(14)))  S(a)+S(−a)=((12)/(1−a))+((12)/(1+a))=((24)/(1−a^2 ))  =24×14=336

$$\mathrm{S}\left(\mathrm{r}\right)=\frac{\mathrm{12}}{\mathrm{1}−\mathrm{r}} \\ $$ $$\mathrm{S}\left(\mathrm{a}\right)=\frac{\mathrm{12}}{\mathrm{1}−\mathrm{a}} \\ $$ $$\mathrm{S}\left(−\mathrm{a}\right)=\frac{\mathrm{12}}{\mathrm{1}+\mathrm{a}} \\ $$ $$\mathrm{S}\left(\mathrm{a}\right)\centerdot\mathrm{S}\left(−\mathrm{a}\right)=\frac{\mathrm{12}}{\mathrm{1}−\mathrm{a}}×\frac{\mathrm{12}}{\mathrm{1}+\mathrm{a}}=\mathrm{2016} \\ $$ $$\frac{\mathrm{144}}{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }=\mathrm{2016} \\ $$ $$\frac{\mathrm{1}}{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }=\frac{\mathrm{2016}}{\mathrm{144}}=\mathrm{14} \\ $$ $$\mathrm{a}=\pm\sqrt{\frac{\mathrm{13}}{\mathrm{14}}} \\ $$ $$\mathrm{S}\left(\mathrm{a}\right)+\mathrm{S}\left(−\mathrm{a}\right)=\frac{\mathrm{12}}{\mathrm{1}−\mathrm{a}}+\frac{\mathrm{12}}{\mathrm{1}+\mathrm{a}}=\frac{\mathrm{24}}{\mathrm{1}−\mathrm{a}^{\mathrm{2}} } \\ $$ $$=\mathrm{24}×\mathrm{14}=\mathrm{336} \\ $$

Commented byJoel575 last updated on 11/Dec/16

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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