Question-183750

Question Number 183750 by kapoorshah last updated on 29/Dec/22

Answered by mr W last updated on 30/Dec/22

S_1 =a S_6 =((a(1−q^6 ))/(1−q)) a+((a(1−q^6 ))/(1−q))=1024 a(2−q−q^6 )=1024(1−q) ((a(1−q^3 ))/(1−q))×((a(1−q^4 ))/(1−q))=1023 ((a^2 (1−q^3 )(1−q^4 ))/((1−q)^2 ))=1023 ((1024^2 (1−q^3 )(1−q^4 ))/((2−q−q^6 )^2 ))=1023 ⇒q≈3.9841 (S_(11) /S_8 )=((1−q^(11) )/(1−q^8 ))≈63.2408

$${S}_{\mathrm{1}} ={a} \\ $$$${S}_{\mathrm{6}} =\frac{{a}\left(\mathrm{1}−{q}^{\mathrm{6}} \right)}{\mathrm{1}−{q}} \\ $$$${a}+\frac{{a}\left(\mathrm{1}−{q}^{\mathrm{6}} \right)}{\mathrm{1}−{q}}=\mathrm{1024} \\ $$$${a}\left(\mathrm{2}−{q}−{q}^{\mathrm{6}} \right)=\mathrm{1024}\left(\mathrm{1}−{q}\right) \\ $$$$\frac{{a}\left(\mathrm{1}−{q}^{\mathrm{3}} \right)}{\mathrm{1}−{q}}×\frac{{a}\left(\mathrm{1}−{q}^{\mathrm{4}} \right)}{\mathrm{1}−{q}}=\mathrm{1023} \\ $$$$\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−{q}^{\mathrm{3}} \right)\left(\mathrm{1}−{q}^{\mathrm{4}} \right)}{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} }=\mathrm{1023} \\ $$$$\frac{\mathrm{1024}^{\mathrm{2}} \left(\mathrm{1}−{q}^{\mathrm{3}} \right)\left(\mathrm{1}−{q}^{\mathrm{4}} \right)}{\left(\mathrm{2}−{q}−{q}^{\mathrm{6}} \right)^{\mathrm{2}} }=\mathrm{1023} \\ $$$$\Rightarrow{q}\approx\mathrm{3}.\mathrm{9841} \\ $$$$\frac{{S}_{\mathrm{11}} }{{S}_{\mathrm{8}} }=\frac{\mathrm{1}−{q}^{\mathrm{11}} }{\mathrm{1}−{q}^{\mathrm{8}} }\approx\mathrm{63}.\mathrm{2408} \\ $$

Commented by kapoorshah last updated on 30/Dec/22