Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

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 2S the sum to infinity of the cubes of the terms of the series is ((64)/(13))S. find the value of S and write iut the first 3 terms if the series.

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

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

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

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

thank you

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

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

please help Qn 77990

$${please}\:{help}\:{Qn}\:\mathrm{77990} \\ $$

Commented by Rio Michael last updated on 18/Jan/20

thank you sirs

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com