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Question-186805




Question Number 186805 by pascal889 last updated on 10/Feb/23
Answered by Rasheed.Sindhi last updated on 10/Feb/23
r=(a_4 /a_(10) )=(a_1 /a_4 )  r=((a+3d)/(a+9d))=(a/(a+3d))  r−1=((a+3d)/(a+9d))−1=(a/(a+3d))−1  r−1=((−6d)/(a+9d))=((−3d)/(a+3d))  ((r−1)/(−3d))=(2/(a+9d))=(1/(a+3d))              a+9d=2a+6d             a=3d  ((r−1)/(−3d))=(1/(a+3d))⇒((r−1)/(−a))=(1/(2a))              1−r=(1/2)               r=(1/2)  first term of gp: 4  common ratio=(1/2)  S_6 =4(((1−(1/2)^6 )/(1−1/2)))=4(((63/64)/(1/2)))             4((((63)/(64))/(1/2)))=4×((63)/(64))×2=((63)/8)=7(7/8)
$${r}=\frac{{a}_{\mathrm{4}} }{{a}_{\mathrm{10}} }=\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{4}} } \\ $$$${r}=\frac{{a}+\mathrm{3}{d}}{{a}+\mathrm{9}{d}}=\frac{{a}}{{a}+\mathrm{3}{d}} \\ $$$${r}−\mathrm{1}=\frac{{a}+\mathrm{3}{d}}{{a}+\mathrm{9}{d}}−\mathrm{1}=\frac{{a}}{{a}+\mathrm{3}{d}}−\mathrm{1} \\ $$$${r}−\mathrm{1}=\frac{−\mathrm{6}{d}}{{a}+\mathrm{9}{d}}=\frac{−\mathrm{3}{d}}{{a}+\mathrm{3}{d}} \\ $$$$\frac{{r}−\mathrm{1}}{−\mathrm{3}{d}}=\frac{\mathrm{2}}{{a}+\mathrm{9}{d}}=\frac{\mathrm{1}}{{a}+\mathrm{3}{d}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{a}+\mathrm{9}{d}=\mathrm{2}{a}+\mathrm{6}{d} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{a}=\mathrm{3}{d} \\ $$$$\frac{{r}−\mathrm{1}}{−\mathrm{3}{d}}=\frac{\mathrm{1}}{{a}+\mathrm{3}{d}}\Rightarrow\frac{{r}−\mathrm{1}}{−{a}}=\frac{\mathrm{1}}{\mathrm{2}{a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−{r}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{r}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${first}\:{term}\:{of}\:{gp}:\:\mathrm{4} \\ $$$${common}\:{ratio}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${S}_{\mathrm{6}} =\mathrm{4}\left(\frac{\mathrm{1}−\left(\mathrm{1}/\mathrm{2}\right)^{\mathrm{6}} }{\mathrm{1}−\mathrm{1}/\mathrm{2}}\right)=\mathrm{4}\left(\frac{\mathrm{63}/\mathrm{64}}{\mathrm{1}/\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\left(\frac{\frac{\mathrm{63}}{\mathrm{64}}}{\frac{\mathrm{1}}{\mathrm{2}}}\right)=\mathrm{4}×\frac{\mathrm{63}}{\mathrm{64}}×\mathrm{2}=\frac{\mathrm{63}}{\mathrm{8}}=\mathrm{7}\frac{\mathrm{7}}{\mathrm{8}} \\ $$

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