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Question Number 79649 by john santu last updated on 27/Jan/20

$$\mathrm{given}\mathrm{a},\mathrm{ar},{\mathrm{ar}}^2,{\mathrm{ar}}^3,...\mathrm{is}\mathrm{a}\mathrm{GPwith}$$ $$\mathrm{n}\to \mathrm{\infty },\mathrm{r}<1$$ $$\mathrm{if}:\mathrm{a},{\mathrm{x}}_1,{\mathrm{x}}_2,\mathrm{ar},{\mathrm{x}}_3,{\mathrm{x}}_4,{\mathrm{x}}_5,{\mathrm{x}}_6,{\mathrm{ar}}^2,$$ $${\mathrm{x}}_7,{\mathrm{x}}_8,{\mathrm{x}}_9,{\mathrm{x}}_{10},{\mathrm{x}}_{11},{\mathrm{x}}_{12},{\mathrm{ar}}^3,....$$ $$\mathrm{where}:\mathrm{a},{\mathrm{x}}_1,{\mathrm{x}}_2,\mathrm{ar}\Rightarrow \mathrm{AP}$$ $$\mathrm{ar},{\mathrm{x}}_3,{\mathrm{x}}_4,{\mathrm{x}}_5,{\mathrm{x}}_6,{\mathrm{ar}}^2\Rightarrow \mathrm{AP}$$ $${\mathrm{ar}}^2,{\mathrm{x}}_7,{\mathrm{x}}_8,{\mathrm{x}}_9,{\mathrm{x}}_{10},{\mathrm{x}}_{11},{\mathrm{x}}_{12},{\mathrm{ar}}^3\Rightarrow \mathrm{AP}$$ $$...\mathrm{etc}$$ $$\mathrm{if}\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }({\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+...)=\frac{21}{16}\times \frac{\mathrm{a}}{1-\mathrm{r}}$$ $$\mathrm{what}\mathrm{is}\mathrm{r}?$$

Commented byjohn santu last updated on 27/Jan/20

$$\mathrm{mister}\mathrm{Mjs},\mathrm{W},\mathrm{mind}\mathrm{is}\mathrm{power}\mathrm{i}\mathrm{need}\mathrm{your}\mathrm{help}$$ $$$$

Commented bymind is power last updated on 27/Jan/20

$$a,x_1,x_2,arAp$$ $$s=2a(1+r)$$ $$x_1+x_2=2a(1+r)-a(1+r)=a(1+r)$$ $$x_3+x_4+x_5+x_6=2(ar+{ar}^2)$$ $$x_7+........+x_{12}=3({ar}^2+{ar}^4)$$ $$weget\mathrm{\Sigma }{x}_k=\mathrm{\Sigma }(a+2ar+3{ar}^2+.......)+\mathrm{\Sigma }(ar+2{ar}^2+........)$$ $$=a\mathrm{\Sigma }(1+2r+3r^2+....)+ar\mathrm{\Sigma }(1+2r+3r^2+.......)$$ $$=(a+ar)\mathrm{\Sigma }(1+2r+3r^2+....)$$ $$\sum _{k⩾1}{kr}^{k-1}=\mathrm{\Sigma }\frac{d}{dr}\left(r^k\right)=\frac{d}{dr}\mathrm{\Sigma }{r}^k=\frac{d}{dr}\left(\frac{r}{1-r}\right)=\frac{1}{{(1-r)}^2}$$ $$\Rightarrow \frac{a(1+r)}{{(1-r)}^2}=\frac{21a}{16(1-r)},r\in ]0,1[$$ $$\Rightarrow 21(1-r)=16(1+r)\Rightarrow 5=37r\Rightarrow r=\frac{5}{37}$$ $$$$

Commented byjohn santu last updated on 27/Jan/20

$$\mathrm{sir}\mathrm{the}\mathrm{option}$$ $$\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}\mathrm{and}\frac{1}{6}$$

Commented byjohn santu last updated on 27/Jan/20

$$\mathrm{may}\mathrm{be}\mathrm{it}\mathrm{answer}\mathrm{in}\mathrm{my}\mathrm{book}\mathrm{wrong}$$

Commented byjohn santu last updated on 27/Jan/20

$$\mathrm{thank}\mathrm{you}\mathrm{mr}\mathrm{Mind}\mathrm{is}\mathrm{power},\mathrm{W}$$

Commented bymind is power last updated on 27/Jan/20

$$withepleasur$$

Answered by mr W last updated on 27/Jan/20

$${ar}^k,x_{m+1},x_{m+2},...,x_{m+2(k+1)},{ar}^{k+1}areAP$$ $$withm=\underset{j=0}{\sum ^{k-1}}2(j+1)=2\times \frac{k(k+1)}{2}=k(k+1)$$ $$saythisAPis:$$ $$b_0,b_1,...,b_{2(k+1)},b_{2(k+1)+1}$$ $$\mathrm{\Sigma }b=\frac{(b_0+b_{2(k+1)+1})(2(k+1)+2)}{2}=(k+2)({ar}^k+{ar}^{k+1})$$ $$\mathrm{\Sigma }b={ar}^k+\mathrm{\Sigma }x+{ar}^{k+1}=(k+2)({ar}^k+{ar}^{k+1})$$ $$\Rightarrow \mathrm{\Sigma }x=(k+1)({ar}^k+{ar}^{k+1})=a(1+r)(k+1)r^k$$ $$\Rightarrow \sum _{all}x=\underset{k=0}{\sum ^{\mathrm{\infty }}}a(1+r)(k+1)r^k$$ $$\Rightarrow \sum _{all}x=\frac{a(1+r)}{r}\underset{k=1}{\sum ^{\mathrm{\infty }}}{kr}^k$$ $$\Rightarrow \sum _{all}x=\frac{a(1+r)}{r}S$$ $$S=1r+2r^2+3r^3+...$$ $$rS=1r^2+2r^3+3r^4+...$$ $$S-rS=r+r^2+r^3+...=\frac{r}{1-r}$$ $$(1-r)S=\frac{r}{1-r}$$ $$S=\frac{r}{{(1-r)}^2}$$ $$\Rightarrow \sum _{all}x=\frac{a(1+r)}{r}\times \frac{r}{{(1-r)}^2}=\frac{a(1+r)}{{(1-r)}^2}$$ $$\Rightarrow \frac{a(1+r)}{{(1-r)}^2}=\frac{21}{16}\times \frac{a}{1-r}$$ $$\Rightarrow \frac{1+r}{1-r}=\frac{21}{16}$$ $$\Rightarrow 16+16r=21-21r$$ $$\Rightarrow 37r=5$$ $$\Rightarrow r=\frac{5}{37}$$

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