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

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