the-sum-to-infinity-of-a-Geometric-series-is-S-the-sum-to-infinty-of-the-squares-of-the-terms-of-the-series-is-2S-the-sum-to-infinity-of-the-cubes-of-the-terms-of-the-series-is-64-13-S-find-the-

Question Number 78493 by Rio Michael last updated on 18/Jan/20

$${the}\:{sum}\:{to}\:{infinity}\:{of}\:{a}\:{Geometric}\:{series}\:{is}\:{S} \\ $$$${the}\:{sum}\:{to}\:{infinty}\:{of}\:{the}\:{squares}\:{of}\:{the}\:{terms} \\ $$$${of}\:{the}\:{series}\:{is}\:\mathrm{2}{S} \\ $$$${the}\:{sum}\:{to}\:{infinity}\:{of}\:{the}\:{cubes}\:{of}\:{the}\:{terms} \\ $$$${of}\:{the}\:{series}\:{is}\:\frac{\mathrm{64}}{\mathrm{13}}{S}. \\ $$$${find}\:{the}\:{value}\:{of}\:{S}\:{and}\:{write}\:{iut}\:{the}\:{first} \\ $$$$\mathrm{3}\:{terms}\:{if}\:{the}\:{series}. \\ $$

Answered by mr W last updated on 18/Jan/20

S=(a_1 /(1−q)) ...(i) 2S=(a_1 ^2 /(1−q^2 ))>0 ...(ii) ((64)/(13))S=(a_1 ^3 /(1−q^3 )) ...(iii) a_1 =(1−q)S 2S=(a_1 ^2 /(1−q^2 ))=(((1−q)^2 S^2 )/(1−q^2 ))=((1−q)/(1+q))S^2 2=((1−q)/(1+q))S ⇒q=((S−2)/(S+2))=1−(4/(S+2)) ⇒a_1 =((4S)/(S+2)) ((64)/(13))S=(a_1 ^3 /(1−q^3 ))=(((((4S)/(S+2)))^3 )/(1−(((S−2)/(S+2)))^3 )) ((64)/(13))S=((4^3 S^3 )/((S+2)^3 −(S−2)^3 ))=((64S^3 )/(2(6S^2 +8))) (1/(13))=(S^2 /(2(6S^2 +8))) S^2 =16 ⇒S=4 ⇒q=((S−2)/(S+2))=((4−2)/(4+2))=(1/3) ⇒a_1 =((4S)/(S+2))=((16)/6)=(8/3) ⇒a_2 =(8/3)×(1/3)=(8/9) ⇒a_3 =(8/9)×(1/3)=(8/(27))

$${S}=\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{q}}\:\:\:…\left({i}\right) \\ $$$$\mathrm{2}{S}=\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{1}−{q}^{\mathrm{2}} }>\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\frac{\mathrm{64}}{\mathrm{13}}{S}=\frac{{a}_{\mathrm{1}} ^{\mathrm{3}} }{\mathrm{1}−{q}^{\mathrm{3}} }\:\:\:…\left({iii}\right) \\ $$$${a}_{\mathrm{1}} =\left(\mathrm{1}−{q}\right){S} \\ $$$$\mathrm{2}{S}=\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{1}−{q}^{\mathrm{2}} }=\frac{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} {S}^{\mathrm{2}} }{\mathrm{1}−{q}^{\mathrm{2}} }=\frac{\mathrm{1}−{q}}{\mathrm{1}+{q}}{S}^{\mathrm{2}} \\ $$$$\mathrm{2}=\frac{\mathrm{1}−{q}}{\mathrm{1}+{q}}{S} \\ $$$$\Rightarrow{q}=\frac{{S}−\mathrm{2}}{{S}+\mathrm{2}}=\mathrm{1}−\frac{\mathrm{4}}{{S}+\mathrm{2}} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\frac{\mathrm{4}{S}}{{S}+\mathrm{2}} \\ $$$$ \\ $$$$\frac{\mathrm{64}}{\mathrm{13}}{S}=\frac{{a}_{\mathrm{1}} ^{\mathrm{3}} }{\mathrm{1}−{q}^{\mathrm{3}} }=\frac{\left(\frac{\mathrm{4}{S}}{{S}+\mathrm{2}}\right)^{\mathrm{3}} }{\mathrm{1}−\left(\frac{{S}−\mathrm{2}}{{S}+\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$\frac{\mathrm{64}}{\mathrm{13}}{S}=\frac{\mathrm{4}^{\mathrm{3}} {S}^{\mathrm{3}} }{\left({S}+\mathrm{2}\right)^{\mathrm{3}} −\left({S}−\mathrm{2}\right)^{\mathrm{3}} }=\frac{\mathrm{64}{S}^{\mathrm{3}} }{\mathrm{2}\left(\mathrm{6}{S}^{\mathrm{2}} +\mathrm{8}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{13}}=\frac{{S}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{6}{S}^{\mathrm{2}} +\mathrm{8}\right)} \\ $$$${S}^{\mathrm{2}} =\mathrm{16} \\ $$$$\Rightarrow{S}=\mathrm{4} \\ $$$$\Rightarrow{q}=\frac{{S}−\mathrm{2}}{{S}+\mathrm{2}}=\frac{\mathrm{4}−\mathrm{2}}{\mathrm{4}+\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\frac{\mathrm{4}{S}}{{S}+\mathrm{2}}=\frac{\mathrm{16}}{\mathrm{6}}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\Rightarrow{a}_{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{8}}{\mathrm{9}} \\ $$$$\Rightarrow{a}_{\mathrm{3}} =\frac{\mathrm{8}}{\mathrm{9}}×\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{8}}{\mathrm{27}} \\ $$

Commented by peter frank last updated on 18/Jan/20

$${thank}\:{you} \\ $$

Answered by john santu last updated on 18/Jan/20

a+ar+ar^2 +ar^3 +...=(a/(1−r))=S ⇒a=S(1−r) a^2 +a^2 r^2 +a^2 r^4 +a^2 r^6 +...=(a^2 /(1−r^2 )) =2S (a/(1+r))×(a/(1−r))=2S (a/(1+r))=2 ⇒a=2+2r ⇒S−Sr=2+2r a^3 (1+r^3 +r^6 +r^9 +...)=((64)/(13))S (a^3 /(1−r^3 ))=((64)/(13))S ⇒(a/((1−r)))×(a^2 /((1+r+r^2 )))=((64)/(13))S 13a^2 =64(1+r+r^2 ) 13×4(r^2 +2r+1)=64(r^2 +r+1) 13r^2 +26r+13= 16r^2 +16r+16 3r^2 −10r+3=0⇒(3r−1)(r−3)=0 r=3 (no solution) r=(1/3) ⇒ a=2+(2/3)=(8/3) S= (((8/3))/(1−(1/3)))= 4

$${a}+{ar}+{ar}^{\mathrm{2}} +{ar}^{\mathrm{3}} +…=\frac{{a}}{\mathrm{1}−{r}}={S}\:\Rightarrow{a}={S}\left(\mathrm{1}−{r}\right) \\ $$$${a}^{\mathrm{2}} +{a}^{\mathrm{2}} {r}^{\mathrm{2}} +{a}^{\mathrm{2}} {r}^{\mathrm{4}} +{a}^{\mathrm{2}} {r}^{\mathrm{6}} +…=\frac{{a}^{\mathrm{2}} }{\mathrm{1}−{r}^{\mathrm{2}} }\:=\mathrm{2}{S} \\ $$$$\frac{{a}}{\mathrm{1}+{r}}×\frac{{a}}{\mathrm{1}−{r}}=\mathrm{2}{S} \\ $$$$\frac{{a}}{\mathrm{1}+{r}}=\mathrm{2}\:\Rightarrow{a}=\mathrm{2}+\mathrm{2}{r}\:\Rightarrow{S}−{Sr}=\mathrm{2}+\mathrm{2}{r} \\ $$$${a}^{\mathrm{3}} \left(\mathrm{1}+{r}^{\mathrm{3}} +{r}^{\mathrm{6}} +{r}^{\mathrm{9}} +…\right)=\frac{\mathrm{64}}{\mathrm{13}}{S} \\ $$$$\frac{{a}^{\mathrm{3}} }{\mathrm{1}−{r}^{\mathrm{3}} }=\frac{\mathrm{64}}{\mathrm{13}}{S}\:\:\Rightarrow\frac{{a}}{\left(\mathrm{1}−{r}\right)}×\frac{{a}^{\mathrm{2}} }{\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)}=\frac{\mathrm{64}}{\mathrm{13}}{S} \\ $$$$\mathrm{13}{a}^{\mathrm{2}} =\mathrm{64}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right) \\ $$$$\mathrm{13}×\mathrm{4}\left({r}^{\mathrm{2}} +\mathrm{2}{r}+\mathrm{1}\right)=\mathrm{64}\left({r}^{\mathrm{2}} +{r}+\mathrm{1}\right) \\ $$$$\mathrm{13}{r}^{\mathrm{2}} +\mathrm{26}{r}+\mathrm{13}=\:\mathrm{16}{r}^{\mathrm{2}} +\mathrm{16}{r}+\mathrm{16} \\ $$$$\mathrm{3}{r}^{\mathrm{2}} −\mathrm{10}{r}+\mathrm{3}=\mathrm{0}\Rightarrow\left(\mathrm{3}{r}−\mathrm{1}\right)\left({r}−\mathrm{3}\right)=\mathrm{0} \\ $$$${r}=\mathrm{3}\:\left({no}\:{solution}\right) \\ $$$${r}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{a}=\mathrm{2}+\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$${S}=\:\frac{\left(\mathrm{8}/\mathrm{3}\right)}{\mathrm{1}−\left(\mathrm{1}/\mathrm{3}\right)}=\:\mathrm{4} \\ $$

Commented by peter frank last updated on 18/Jan/20