Question Number 101779 by I want to learn more last updated on 04/Jul/20
$$\mathrm{If}\:\:\:\:\mathrm{S}_{\mathrm{n}} \:\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\:\mathrm{n}\:\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}. \\ $$$$\mathrm{Express}\:\:\:\mathrm{S}_{\mathrm{2k}} \:\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\:\:\mathrm{S}_{\mathrm{k}} \:\:\mathrm{and}\:\:\:\mathrm{S}_{\mathrm{3k}} \\ $$
Answered by bemath last updated on 04/Jul/20
$${S}_{\mathrm{n}} \:=\:\frac{\mathrm{n}\left(\mathrm{a}+\mathrm{u}_{\mathrm{n}} \right)}{\mathrm{2}}\rightarrow\:\mathrm{S}_{{k}\:} =\:\frac{{k}\left({a}+{u}_{{k}} \right)}{\mathrm{2}} \\ $$$${u}_{{k}} =\:\frac{\mathrm{2}{S}_{{k}} }{{k}}−{a}\:\Rightarrow{a}+\left({k}−\mathrm{1}\right){d}=\frac{\mathrm{2}{S}_{{k}} }{{k}}−{a} \\ $$$$\left({k}−\mathrm{1}\right){d}\:=\:\frac{\mathrm{2}{S}_{{k}} }{{k}}−\mathrm{2}{a}\Rightarrow{d}\:=\frac{\mathrm{2}{S}_{{k}} −\mathrm{2}{ak}}{{k}\left({k}−\mathrm{1}\right)}\:\left(\bullet\right) \\ $$$${S}_{\mathrm{2}{k}} \:=\:\frac{\mathrm{2}{k}\left({a}+{u}_{\mathrm{2}{k}} \right)}{\mathrm{2}}\:=\:{k}\left({a}+{u}_{\mathrm{2}{k}} \right) \\ $$$${u}_{\mathrm{2}{k}} \:=\:\frac{{S}_{\mathrm{2}{k}} }{{k}}−{a}\:\Leftrightarrow{a}+\left(\mathrm{2}{k}−\mathrm{1}\right){d}=\frac{{S}_{\mathrm{2}{k}} }{{k}}−{a}\: \\ $$$${d}\:=\:\frac{{S}_{\mathrm{2}{k}} −{ak}}{{k}\left(\mathrm{2}{k}−\mathrm{1}\right)}\:\left(\bullet\bullet\right) \\ $$$$\left(\bullet\right)=\left(\bullet\bullet\right) \\ $$$$\frac{\mathrm{2S}_{{k}} −\mathrm{2}{ak}}{{k}\left({k}−\mathrm{1}\right)}\:=\:\frac{{S}_{\mathrm{2}{k}} −{ak}}{{k}\left(\mathrm{2}{k}−\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{S}_{{k}} −\mathrm{2}{ak}\right)}{{k}−\mathrm{1}}\:+{ak}\:=\:{S}_{\mathrm{2}{k}} \: \\ $$$$ \\ $$
Commented by I want to learn more last updated on 05/Jul/20
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$