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JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    In a geometric series, the first term \[=a,\] common ratio\[=r\]. If \[{{S}_{n}}\] denotes the sum of the \[n\] terms and\[{{U}_{n}}=\sum\limits_{n=1}^{n}{{{S}_{n}}}\], then\[r{{S}_{n}}+(1-r){{U}_{n}}\] equals to

    A) \[0\]                                     

    B) \[n\]

    C) \[na\]                                   

    D) \[nar\]

    Correct Answer: B

    Solution :

    \[{{S}_{n}}=\frac{a({{r}^{n}}-1)}{r-1}\] \[\Rightarrow \]               \[{{U}_{n}}=\sum\limits_{n=1}^{n}{{}}\frac{a({{r}^{n}}-1)}{r-1}=\frac{a}{r-1}\sum\limits_{n=1}^{n}{({{r}^{n}}-1)}\] \[\Rightarrow \]               \[{{U}_{n}}=\frac{a}{x-1}\{r+{{r}^{2}}+...+{{r}^{n}}-n\}\]                 \[=\frac{a}{r-1}\left\{ \frac{r({{r}^{n}}-1)}{r-1}-n \right\}\] \[\Rightarrow \]               \[(r-1){{U}_{n}}=\frac{ar({{r}^{n}}-1)}{r-1}-an\] \[\Rightarrow \]               \[r{{S}_{n}}+(1-r){{U}_{n}}=an\]


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