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-

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

$$thesumtoinfinityofaGeometricseriesisS$$$$thesumtoinfintyofthesquaresoftheterms$$$$oftheseriesis2S$$$$thesumtoinfinityofthecubesoftheterms$$$$oftheseriesis\frac{64}{13}S.$$$$findthevalueofSandwriteiutthefirst$$$$3termsiftheseries.$$

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

$$S=\frac{a_1}{1-q}\dots \left(i\right)$$$$2S=\frac{a_1^2}{1-q^2}>0\dots \left(ii\right)$$$$\frac{64}{13}S=\frac{a_1^3}{1-q^3}\dots \left(iii\right)$$$$a_1=(1-q)S$$$$2S=\frac{a_1^2}{1-q^2}=\frac{{(1-q)}^2{S}^2}{1-q^2}=\frac{1-q}{1+q}{S}^2$$$$2=\frac{1-q}{1+q}S$$$$\Rightarrow q=\frac{S-2}{S+2}=1-\frac{4}{S+2}$$$$\Rightarrow a_1=\frac{4S}{S+2}$$$$$$$$\frac{64}{13}S=\frac{a_1^3}{1-q^3}=\frac{{\left(\frac{4S}{S+2}\right)}^3}{1-{\left(\frac{S-2}{S+2}\right)}^3}$$$$\frac{64}{13}S=\frac{4^3{S}^3}{{(S+2)}^3-{(S-2)}^3}=\frac{64S^3}{2(6S^2+8)}$$$$\frac{1}{13}=\frac{S^2}{2(6S^2+8)}$$$$S^2=16$$$$\Rightarrow S=4$$$$\Rightarrow q=\frac{S-2}{S+2}=\frac{4-2}{4+2}=\frac{1}{3}$$$$\Rightarrow a_1=\frac{4S}{S+2}=\frac{16}{6}=\frac{8}{3}$$$$\Rightarrow a_2=\frac{8}{3}\times \frac{1}{3}=\frac{8}{9}$$$$\Rightarrow a_3=\frac{8}{9}\times \frac{1}{3}=\frac{8}{27}$$

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

$$thankyou$$

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

$$a+ar+{ar}^2+{ar}^3+\dots =\frac{a}{1-r}=S\Rightarrow a=S(1-r)$$$$a^2+a^2{r}^2+a^2{r}^4+a^2{r}^6+\dots =\frac{a^2}{1-r^2}=2S$$$$\frac{a}{1+r}\times \frac{a}{1-r}=2S$$$$\frac{a}{1+r}=2\Rightarrow a=2+2r\Rightarrow S-Sr=2+2r$$$$a^3(1+r^3+r^6+r^9+\dots )=\frac{64}{13}S$$$$\frac{a^3}{1-r^3}=\frac{64}{13}S\Rightarrow \frac{a}{(1-r)}\times \frac{a^2}{(1+r+r^2)}=\frac{64}{13}S$$$$13a^2=64(1+r+r^2)$$$$13\times 4(r^2+2r+1)=64(r^2+r+1)$$$$13r^2+26r+13=16r^2+16r+16$$$$3r^2-10r+3=0\Rightarrow (3r-1)(r-3)=0$$$$r=3\left(nosolution\right)$$$$r=\frac{1}{3}\Rightarrow a=2+\frac{2}{3}=\frac{8}{3}$$$$S=\frac{\left(8/3\right)}{1-\left(1/3\right)}=4$$

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

$$thankyou$$

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

$$pleasehelpQn77990$$

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

$$thankyousirs$$

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