Question Number 46918 by 786786AM last updated on 02/Nov/18
$$\mathrm{Let}\:{S}_{{n}} \:\mathrm{denote}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:{n}\:\mathrm{terms}\:\mathrm{of} \\ $$$$\mathrm{an}\:\mathrm{AP}.\:\mathrm{If}\:\:\:{S}_{\mathrm{2}{n}} =\:\mathrm{3}\:{S}_{{n}} \:,\:\mathrm{then}\:\mathrm{the}\:\mathrm{ratio} \\ $$$$\frac{{S}_{\mathrm{3}{n}} }{{S}_{{n}} }\:\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18
![(s_(2n) /s_n )=((((2n)/2)[2a+(2n−1)d])/((n/2)[2a+(n−1)d]))=3 4a+(4n−2)d=6a+(3n−3)d d(4n−2−3n+3)=2a a=(((n+1)d)/2) (s_(3n) /s_n )=((((3n)/2)[2a+(3n−1)d])/((n/2)[2a+(n−1)d])) =((3[(n+1)d+(3n−1)d])/([(n+1)d+(n−1)d])) =3×((d(n+1+3n−1))/(d(n+1+n−1))) =3×((4n)/(2n))=6](https://www.tinkutara.com/question/Q46937.png)
$$\frac{{s}_{\mathrm{2}{n}} }{{s}_{{n}} }=\frac{\frac{\mathrm{2}{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{2}{n}−\mathrm{1}\right){d}\right]}{\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right]}=\mathrm{3} \\ $$$$\mathrm{4}{a}+\left(\mathrm{4}{n}−\mathrm{2}\right){d}=\mathrm{6}{a}+\left(\mathrm{3}{n}−\mathrm{3}\right){d} \\ $$$${d}\left(\mathrm{4}{n}−\mathrm{2}−\mathrm{3}{n}+\mathrm{3}\right)=\mathrm{2}{a} \\ $$$${a}=\frac{\left({n}+\mathrm{1}\right){d}}{\mathrm{2}} \\ $$$$\frac{{s}_{\mathrm{3}{n}} }{{s}_{{n}} }=\frac{\frac{\mathrm{3}{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{3}{n}−\mathrm{1}\right){d}\right]}{\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right]} \\ $$$$=\frac{\mathrm{3}\left[\left({n}+\mathrm{1}\right){d}+\left(\mathrm{3}{n}−\mathrm{1}\right){d}\right]}{\left[\left({n}+\mathrm{1}\right){d}+\left({n}−\mathrm{1}\right){d}\right]} \\ $$$$=\mathrm{3}×\frac{{d}\left({n}+\mathrm{1}+\mathrm{3}{n}−\mathrm{1}\right)}{{d}\left({n}+\mathrm{1}+{n}−\mathrm{1}\right)} \\ $$$$=\mathrm{3}×\frac{\mathrm{4}{n}}{\mathrm{2}{n}}=\mathrm{6} \\ $$