Question-22939

Question Number 22939 by selestian last updated on 24/Oct/17

Answered by ajfour last updated on 24/Oct/17

(B) 1/8 (a/(1−r))=4 ; (a^3 /(1−r^3 ))=((64)/7) ⇒ ((7a^3 )/(1−r^3 ))=(a^3 /((1−r)^3 )) ⇒ 7(1−r)^2 =1+r+r^2 7+7r^2 −14r=1+r+r^2 2r^2 −5r+2=0 (2r−1)(r−2)=0 ⇒ r=(1/2) for the decreasing G.P. (a/(1−r))=4 ⇒ a=4(1−(1/2)) a=2 , then 5^(th) term ar^4 =(1/8) .

$$\left({B}\right)\:\mathrm{1}/\mathrm{8} \\ $$$$\frac{{a}}{\mathrm{1}−{r}}=\mathrm{4}\:\:\:\:;\:\:\frac{{a}^{\mathrm{3}} }{\mathrm{1}−{r}^{\mathrm{3}} }=\frac{\mathrm{64}}{\mathrm{7}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{7}{a}^{\mathrm{3}} }{\mathrm{1}−{r}^{\mathrm{3}} }=\frac{{a}^{\mathrm{3}} }{\left(\mathrm{1}−{r}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\:\:\mathrm{7}\left(\mathrm{1}−{r}\right)^{\mathrm{2}} =\mathrm{1}+{r}+{r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{7}+\mathrm{7}{r}^{\mathrm{2}} −\mathrm{14}{r}=\mathrm{1}+{r}+{r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{r}^{\mathrm{2}} −\mathrm{5}{r}+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{2}{r}−\mathrm{1}\right)\left({r}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{r}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{for}\:{the}\:{decreasing}\:{G}.{P}. \\ $$$$\:\:\:\:\:\:\:\:\frac{{a}}{\mathrm{1}−{r}}=\mathrm{4}\:\:\:\Rightarrow\:\:{a}=\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:{a}=\mathrm{2}\:\:,\:{then}\:\mathrm{5}^{{th}} \:{term}\:{ar}^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{8}}\:. \\ $$

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Question-22939

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