Question Number 183750 by kapoorshah last updated on 29/Dec/22
Answered by mr W last updated on 30/Dec/22
$${S}_{\mathrm{1}} ={a} \\ $$$${S}_{\mathrm{6}} =\frac{{a}\left(\mathrm{1}−{q}^{\mathrm{6}} \right)}{\mathrm{1}−{q}} \\ $$$${a}+\frac{{a}\left(\mathrm{1}−{q}^{\mathrm{6}} \right)}{\mathrm{1}−{q}}=\mathrm{1024} \\ $$$${a}\left(\mathrm{2}−{q}−{q}^{\mathrm{6}} \right)=\mathrm{1024}\left(\mathrm{1}−{q}\right) \\ $$$$\frac{{a}\left(\mathrm{1}−{q}^{\mathrm{3}} \right)}{\mathrm{1}−{q}}×\frac{{a}\left(\mathrm{1}−{q}^{\mathrm{4}} \right)}{\mathrm{1}−{q}}=\mathrm{1023} \\ $$$$\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−{q}^{\mathrm{3}} \right)\left(\mathrm{1}−{q}^{\mathrm{4}} \right)}{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} }=\mathrm{1023} \\ $$$$\frac{\mathrm{1024}^{\mathrm{2}} \left(\mathrm{1}−{q}^{\mathrm{3}} \right)\left(\mathrm{1}−{q}^{\mathrm{4}} \right)}{\left(\mathrm{2}−{q}−{q}^{\mathrm{6}} \right)^{\mathrm{2}} }=\mathrm{1023} \\ $$$$\Rightarrow{q}\approx\mathrm{3}.\mathrm{9841} \\ $$$$\frac{{S}_{\mathrm{11}} }{{S}_{\mathrm{8}} }=\frac{\mathrm{1}−{q}^{\mathrm{11}} }{\mathrm{1}−{q}^{\mathrm{8}} }\approx\mathrm{63}.\mathrm{2408} \\ $$
Commented by kapoorshah last updated on 30/Dec/22
$${how}\:{to}\:{find}\:{q}\:\approx\:\mathrm{3}.\mathrm{9841} \\ $$
Commented by mr W last updated on 30/Dec/22
$${you}\:{can}\:{only}\:{approximate}. \\ $$$${i}\:{used}\:{an}\:{app}. \\ $$
Commented by kapoorshah last updated on 30/Dec/22
$${what}\:{application}\:{did}\:{you}\:{use}? \\ $$
Commented by mr W last updated on 30/Dec/22
$${i}\:{used}\:{grapher}.\:{but}\:{there}\:{are}\:{many} \\ $$$${other}\:{similar}\:{apps}\:{you}\:{can}\:{also}\:{use}. \\ $$
Commented by kapoorshah last updated on 30/Dec/22
$${thank}\:{you} \\ $$