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Question Number 79649 by john santu last updated on 27/Jan/20
given a,ar,ar^2 ,ar^3 ,... is a GPwith n→∞ ,r < 1 if : a,x_1 ,x_2 ,ar,x_3 , x_4 ,x_5 ,x_6 ,ar^2 , x_7 ,x_8 ,x_9 ,x_(10) ,x_(11) ,x_(12) , ar^3 ,... . where : a,x_1 ,x_2 ,ar ⇒AP ar,x_3 ,x_4 ,x_5 ,x_6 ,ar^2 ⇒AP ar^2 ,x_7 ,x_8 ,x_9 ,x_(10) ,x_(11) ,x_(12) ,ar^3 ⇒AP ...etc if lim_(n→∞) (x_1 +x_2 +x_3 +...)= ((21)/(16))×(a/(1−r)) what is r ?
$$\mathrm{given}\:\mathrm{a},\mathrm{ar},\mathrm{ar}^{\mathrm{2}} ,\mathrm{ar}^{\mathrm{3}} ,…\:\mathrm{is}\:\mathrm{a}\:\mathrm{GPwith}\: \\ $$$$\mathrm{n}\rightarrow\infty\:,\mathrm{r}\:<\:\mathrm{1} \\ $$$$\mathrm{if}\::\:\mathrm{a},\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,\mathrm{ar},\mathrm{x}_{\mathrm{3}} ,\:\mathrm{x}_{\mathrm{4}} ,\mathrm{x}_{\mathrm{5}} ,\mathrm{x}_{\mathrm{6}} ,\mathrm{ar}^{\mathrm{2}} , \\ $$$$\mathrm{x}_{\mathrm{7}} ,\mathrm{x}_{\mathrm{8}} ,\mathrm{x}_{\mathrm{9}} ,\mathrm{x}_{\mathrm{10}} ,\mathrm{x}_{\mathrm{11}} ,\mathrm{x}_{\mathrm{12}} ,\:\mathrm{ar}^{\mathrm{3}} ,…\:. \\ $$$$\mathrm{where}\::\:\mathrm{a},\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,\mathrm{ar}\:\Rightarrow\mathrm{AP} \\ $$$$\mathrm{ar},\mathrm{x}_{\mathrm{3}} ,\mathrm{x}_{\mathrm{4}} ,\mathrm{x}_{\mathrm{5}} ,\mathrm{x}_{\mathrm{6}} ,\mathrm{ar}^{\mathrm{2}} \Rightarrow\mathrm{AP} \\ $$$$\mathrm{ar}^{\mathrm{2}} ,\mathrm{x}_{\mathrm{7}} ,\mathrm{x}_{\mathrm{8}} ,\mathrm{x}_{\mathrm{9}} ,\mathrm{x}_{\mathrm{10}} ,\mathrm{x}_{\mathrm{11}} ,\mathrm{x}_{\mathrm{12}} ,\mathrm{ar}^{\mathrm{3}} \Rightarrow\mathrm{AP} \\ $$$$…\mathrm{etc} \\ $$$$\mathrm{if}\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{3}} +…\right)=\:\frac{\mathrm{21}}{\mathrm{16}}×\frac{\mathrm{a}}{\mathrm{1}−\mathrm{r}} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{r}\:? \\ $$
Commented by john santu last updated on 27/Jan/20
mister Mjs , W, mind is power i need your help
$$\mathrm{mister}\:\mathrm{Mjs}\:,\:\mathrm{W},\:\mathrm{mind}\:\mathrm{is}\:\mathrm{power}\:\mathrm{i}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help} \\ $$$$ \\ $$
Commented by mind is power last updated on 27/Jan/20
a,x_1 ,x_2 ,ar Ap 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 ) we get Σx_k =Σ(a+2ar+3ar^2 +.......)+Σ(ar+2ar^2 +........) =aΣ(1+2r+3r^2 +....)+arΣ(1+2r+3r^2 +.......) =(a+ar)Σ(1+2r+3r^2 +....) Σ_(k≥1) kr^(k−1) =Σ(d/dr)(r^k )=(d/dr)Σr^k =(d/dr)((r/(1−r)))=(1/((1−r)^2 )) ⇒((a(1+r))/((1−r)^2 ))=((21a)/(16(1−r))),r∈]0,1[ ⇒21(1−r)=16(1+r)⇒5=37r⇒r=(5/(37))
$${a},{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,{ar}\:\:\:{Ap} \\ $$$${s}=\mathrm{2}{a}\left(\mathrm{1}+{r}\right) \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\mathrm{2}{a}\left(\mathrm{1}+{r}\right)−{a}\left(\mathrm{1}+{r}\right)={a}\left(\mathrm{1}+{r}\right) \\ $$$${x}_{\mathrm{3}} +{x}_{\mathrm{4}} +{x}_{\mathrm{5}} +{x}_{\mathrm{6}} =\mathrm{2}\left({ar}+{ar}^{\mathrm{2}} \right) \\ $$$${x}_{\mathrm{7}} +……..+{x}_{\mathrm{12}} =\mathrm{3}\left({ar}^{\mathrm{2}} +{ar}^{\mathrm{4}} \right) \\ $$$${we}\:{get}\:\Sigma{x}_{{k}} =\Sigma\left({a}+\mathrm{2}{ar}+\mathrm{3}{ar}^{\mathrm{2}} +…….\right)+\Sigma\left({ar}+\mathrm{2}{ar}^{\mathrm{2}} +……..\right) \\ $$$$={a}\Sigma\left(\mathrm{1}+\mathrm{2}{r}+\mathrm{3}{r}^{\mathrm{2}} +….\right)+{ar}\Sigma\left(\mathrm{1}+\mathrm{2}{r}+\mathrm{3}{r}^{\mathrm{2}} +…….\right) \\ $$$$=\left({a}+{ar}\right)\Sigma\left(\mathrm{1}+\mathrm{2}{r}+\mathrm{3}{r}^{\mathrm{2}} +….\right) \\ $$$$\underset{{k}\geqslant\mathrm{1}} {\sum}{kr}^{{k}−\mathrm{1}} =\Sigma\frac{{d}}{{dr}}\left({r}^{{k}} \right)=\frac{{d}}{{dr}}\Sigma{r}^{{k}} =\frac{{d}}{{dr}}\left(\frac{{r}}{\mathrm{1}−{r}}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} } \\ $$$$\left.\Rightarrow\frac{{a}\left(\mathrm{1}+{r}\right)}{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} }=\frac{\mathrm{21}{a}}{\mathrm{16}\left(\mathrm{1}−{r}\right)},{r}\in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$$\Rightarrow\mathrm{21}\left(\mathrm{1}−{r}\right)=\mathrm{16}\left(\mathrm{1}+{r}\right)\Rightarrow\mathrm{5}=\mathrm{37}{r}\Rightarrow{r}=\frac{\mathrm{5}}{\mathrm{37}} \\ $$$$ \\ $$
Commented by john santu last updated on 27/Jan/20
sir the option (1/2), (1/3),(1/4),(1/5) and (1/6)
$$\mathrm{sir}\:\mathrm{the}\:\mathrm{option}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by john santu last updated on 27/Jan/20
may be it answer in my book wrong
$$\mathrm{may}\:\mathrm{be}\:\mathrm{it}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}\:\mathrm{wrong} \\ $$
Commented by john santu last updated on 27/Jan/20
thank you mr Mind is power, W
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mr}\:\mathrm{Mind}\:\mathrm{is}\:\mathrm{power},\:\mathrm{W} \\ $$
Commented by mind is power last updated on 27/Jan/20
withe pleasur
$${withe}\:{pleasur}\: \\ $$
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) are AP with m=Σ_(j=0) ^(k−1) 2(j+1)=2×((k(k+1))/2)=k(k+1) say this AP is: b_0 ,b_1 ,...,b_(2(k+1)) ,b_(2(k+1)+1) Σb=(((b_0 +b_(2(k+1)+1) )(2(k+1)+2))/2)=(k+2)(ar^k +ar^(k+1) ) Σb=ar^k +Σx+ar^(k+1) =(k+2)(ar^k +ar^(k+1) ) ⇒Σx=(k+1)(ar^k +ar^(k+1) )=a(1+r)(k+1)r^k ⇒Σ_(all) x=Σ_(k=0) ^∞ a(1+r)(k+1)r^k ⇒Σ_(all) x=((a(1+r))/r)Σ_(k=1) ^∞ kr^k ⇒Σ_(all) x=((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 +...=(r/(1−r)) (1−r)S=(r/(1−r)) S=(r/((1−r)^2 )) ⇒Σ_(all) x=((a(1+r))/r)×(r/((1−r)^2 ))=((a(1+r))/((1−r)^2 )) ⇒((a(1+r))/((1−r)^2 ))=((21)/(16))×(a/(1−r)) ⇒((1+r)/(1−r))=((21)/(16)) ⇒16+16r=21−21r ⇒37r=5 ⇒r=(5/(37))
$${ar}^{{k}} ,{x}_{{m}+\mathrm{1}} ,{x}_{{m}+\mathrm{2}} ,…,{x}_{{m}+\mathrm{2}\left({k}+\mathrm{1}\right)} ,{ar}^{{k}+\mathrm{1}} \:{are}\:{AP} \\ $$$${with}\:{m}=\underset{{j}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\mathrm{2}\left({j}+\mathrm{1}\right)=\mathrm{2}×\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}={k}\left({k}+\mathrm{1}\right) \\ $$$${say}\:{this}\:{AP}\:{is}: \\ $$$${b}_{\mathrm{0}} ,{b}_{\mathrm{1}} ,…,{b}_{\mathrm{2}\left({k}+\mathrm{1}\right)} ,{b}_{\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{1}} \\ $$$$\Sigma{b}=\frac{\left({b}_{\mathrm{0}} +{b}_{\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{1}} \right)\left(\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{2}\right)}{\mathrm{2}}=\left({k}+\mathrm{2}\right)\left({ar}^{{k}} +{ar}^{{k}+\mathrm{1}} \right) \\ $$$$\Sigma{b}={ar}^{{k}} +\Sigma{x}+{ar}^{{k}+\mathrm{1}} =\left({k}+\mathrm{2}\right)\left({ar}^{{k}} +{ar}^{{k}+\mathrm{1}} \right) \\ $$$$\Rightarrow\Sigma{x}=\left({k}+\mathrm{1}\right)\left({ar}^{{k}} +{ar}^{{k}+\mathrm{1}} \right)={a}\left(\mathrm{1}+{r}\right)\left({k}+\mathrm{1}\right){r}^{{k}} \\ $$$$\Rightarrow\underset{{all}} {\sum}{x}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{a}\left(\mathrm{1}+{r}\right)\left({k}+\mathrm{1}\right){r}^{{k}} \\ $$$$\Rightarrow\underset{{all}} {\sum}{x}=\frac{{a}\left(\mathrm{1}+{r}\right)}{{r}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{kr}^{{k}} \\ $$$$\Rightarrow\underset{{all}} {\sum}{x}=\frac{{a}\left(\mathrm{1}+{r}\right)}{{r}}{S} \\ $$$${S}=\mathrm{1}{r}+\mathrm{2}{r}^{\mathrm{2}} +\mathrm{3}{r}^{\mathrm{3}} +… \\ $$$${rS}=\mathrm{1}{r}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{3}} +\mathrm{3}{r}^{\mathrm{4}} +… \\ $$$${S}−{rS}={r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} +…=\frac{{r}}{\mathrm{1}−{r}} \\ $$$$\left(\mathrm{1}−{r}\right){S}=\frac{{r}}{\mathrm{1}−{r}} \\ $$$${S}=\frac{{r}}{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{{all}} {\sum}{x}=\frac{{a}\left(\mathrm{1}+{r}\right)}{{r}}×\frac{{r}}{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} }=\frac{{a}\left(\mathrm{1}+{r}\right)}{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{a}\left(\mathrm{1}+{r}\right)}{\left(\mathrm{1}−{r}\right)^{\mathrm{2}} }=\frac{\mathrm{21}}{\mathrm{16}}×\frac{{a}}{\mathrm{1}−{r}} \\ $$$$\Rightarrow\frac{\mathrm{1}+{r}}{\mathrm{1}−{r}}=\frac{\mathrm{21}}{\mathrm{16}} \\ $$$$\Rightarrow\mathrm{16}+\mathrm{16}{r}=\mathrm{21}−\mathrm{21}{r} \\ $$$$\Rightarrow\mathrm{37}{r}=\mathrm{5} \\ $$$$\Rightarrow{r}=\frac{\mathrm{5}}{\mathrm{37}} \\ $$

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