Question Number 38559 by nishant last updated on 27/Jun/18
$${in}\:{a}\:{geometric}\:{series},\:{the}\:{first}\:{term} \\ $$$$={a},\:{common}\:{ratio}={r}.\:{If}\:{S}_{{n}} \:{denotes} \\ $$$${the}\:{sum}\:{of}\:{the}\:{n}\:{terms}\:{and}\:{U}_{{n}} =\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{S}_{{n},} \\ $$$${then}\:{rS}_{{n}} +\left(\mathrm{1}−{r}\right){U}_{{n}\:\:} {equals}\:{to} \\ $$$$\left({a}\right)\:\:\mathrm{0}\:\:\:\:\:\:\left({b}\right)\:\:{n}\:\:\:\:\:\left({c}\right)\:\:\:\:{na}\:\:\:\:\left({d}\right){nar} \\ $$
Answered by MrW3 last updated on 27/Jun/18
$${A}_{{n}} ={ar}^{{n}−\mathrm{1}} \\ $$$${S}_{{n}} ={a}+{ar}+…+{ar}^{{n}−\mathrm{1}} \\ $$$${rS}_{{n}} ={ar}+{ar}^{\mathrm{2}} +…+{ar}^{{n}−\mathrm{1}} +{ar}^{{n}} \\ $$$${rS}_{{n}} ={a}+{ar}+{ar}^{\mathrm{2}} +…+{ar}^{{n}−\mathrm{1}} +{ar}^{{n}} −{a} \\ $$$${rS}_{{n}} ={S}_{{n}} +{ar}^{{n}} −{a} \\ $$$$\Rightarrow{S}_{{n}} =\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}}=\frac{{a}}{\mathrm{1}−{r}}−\frac{{a}}{\mathrm{1}−{r}}×{r}^{{n}} \\ $$$${U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{S}_{{k}} =\frac{{an}}{\mathrm{1}−{r}}−\frac{{a}}{\mathrm{1}−{r}}\left({r}+{r}^{\mathrm{2}} +…+{r}^{{n}} \right) \\ $$$${U}_{{n}} =\frac{{an}}{\mathrm{1}−{r}}−\frac{{a}}{\mathrm{1}−{r}}×\frac{{r}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$$\left(\mathrm{1}−{r}\right){U}_{{n}} ={na}−\frac{{ar}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$${rS}_{{n}} =\frac{{ar}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$$\Rightarrow{rS}_{{n}} +\left(\mathrm{1}−{r}\right){U}_{{n}} ={na} \\ $$$$\Rightarrow{Answer}\:\left({c}\right)\:{is}\:{right}. \\ $$