Question Number 61137 by Tawa1 last updated on 29/May/19
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{3n}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:,\:\mathrm{if}\:\mathrm{the}\:\mathrm{sunm}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{term}\:\mathrm{is} \\ $$$$\mathrm{2n}\:\:\mathrm{and}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{2n}\:\mathrm{term}\:\mathrm{is}\:\:\mathrm{5n} \\ $$
Answered by Kunal12588 last updated on 29/May/19
$${S}_{{n}} =\mathrm{2}{n} \\ $$$$\Rightarrow\frac{{n}}{\mathrm{2}}\left(\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right)=\mathrm{2}{n} \\ $$$$\Rightarrow\mathrm{2}{a}+{nd}−{d}−\mathrm{4}=\mathrm{0} \\ $$$${S}_{\mathrm{2}{n}} =\mathrm{5}{n} \\ $$$$\Rightarrow\frac{\mathrm{2}{n}}{\mathrm{2}}\left(\mathrm{2}{a}+\left(\mathrm{2}{n}−\mathrm{1}\right){d}\right)=\mathrm{5}{n} \\ $$$$\Rightarrow\mathrm{2}{a}+\mathrm{2}{nd}−{d}−\mathrm{5}=\mathrm{0} \\ $$$$ \\ $$$${subtracting}\:{the}\:{red}\:{ones} \\ $$$${nd}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{nd}=\mathrm{1} \\ $$$$ \\ $$$${putting}\:{value}\:{of}\:{nd}\:{in}\:{either}\:{of}\:{red}\:{lines} \\ $$$$\mathrm{2}{a}−{d}−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{a}−{d}=\mathrm{3} \\ $$$$ \\ $$$${S}_{\mathrm{3}{n}} =\frac{\mathrm{3}{n}}{\mathrm{2}}\left(\mathrm{2}{a}+\left(\mathrm{3}{n}−\mathrm{1}\right){d}\right) \\ $$$$=\frac{\mathrm{3}{n}}{\mathrm{2}}\left(\mathrm{2}{a}−{d}+\mathrm{3}{nd}\right) \\ $$$${using}\:{blue}\:{ones} \\ $$$${S}_{\mathrm{3}{n}} =\frac{\mathrm{3}{n}}{\mathrm{2}}\left(\mathrm{3}+\mathrm{3}\right)=\mathrm{9}{n} \\ $$
Commented by Kunal12588 last updated on 29/May/19
$${S}_{{pn}} =\frac{{pn}}{\mathrm{2}}\left(\mathrm{2}{a}+\left({pn}−\mathrm{1}\right){d}\right)=\frac{{p}\left({p}+\mathrm{3}\right){n}}{\mathrm{2}}\:\:;\:{p}\in\mathbb{N} \\ $$
Commented by Tawa1 last updated on 29/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$