Question Number 220738 by Spillover last updated on 18/May/25
Answered by mr W last updated on 18/May/25
$${a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…+{a}_{{n}} +…=\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{q}}={k} \\ $$$${a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +…+{a}_{{n}} ^{\mathrm{2}} +…=\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{1}−{q}^{\mathrm{2}} }={l} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\left(\mathrm{1}−{q}\right){k} \\ $$$$\frac{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} {k}^{\mathrm{2}} }{\mathrm{1}−{q}^{\mathrm{2}} }={l} \\ $$$$\Rightarrow{q}=\frac{{k}^{\mathrm{2}} −{l}}{{k}^{\mathrm{2}} +{l}} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\frac{\mathrm{2}{lk}}{{k}^{\mathrm{2}} +{l}} \\ $$
Commented by Spillover last updated on 18/May/25
$${correct} \\ $$