Question Number 7541 by 314159 last updated on 03/Sep/16
Commented by prakash jain last updated on 03/Sep/16
$${a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} +{r}^{\mathrm{4}} \right)=\mathrm{211} \\ $$$${a}\:{and}\:{r}\:{integers} \\ $$$$\mathrm{211}=\mathrm{1}×\mathrm{211}\Rightarrow{a}=\mathrm{1} \\ $$$$\frac{{r}^{\mathrm{5}} −\mathrm{1}}{{r}−\mathrm{1}}=\mathrm{211} \\ $$
Commented by prakash jain last updated on 03/Sep/16
$$\mathrm{211}\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\:\mathrm{any}\:\mathrm{valid}\:\mathrm{answer}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{question}\:\mathrm{should}\:\mathrm{have}\:\mathrm{been} \\ $$$$\mathrm{sum}=\mathrm{121} \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{9}+\mathrm{27}+\mathrm{81}=\mathrm{121} \\ $$
Commented by prakash jain last updated on 03/Sep/16
$$\mathrm{Let}\:\mathrm{the}\:\mathrm{GP}\:\mathrm{be} \\ $$$$\frac{{a}}{{r}^{\mathrm{2}} },\frac{{a}}{{r}},{a},{ar},{ar}^{\mathrm{2}} \\ $$$$\frac{{a}}{{r}^{\mathrm{2}} }+\frac{{a}}{{r}}+{a}+{ar}+{ar}^{\mathrm{2}} =\mathrm{211} \\ $$$${r}^{\mathrm{2}} +\frac{\mathrm{1}}{{r}^{\mathrm{2}} }+{r}+\frac{\mathrm{1}}{{r}}+\mathrm{1}=\frac{\mathrm{211}}{{a}} \\ $$$$\left({r}+\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}} +\left({r}+\frac{\mathrm{1}}{{r}}\right)−\mathrm{1}=\frac{\mathrm{211}}{{a}} \\ $$$${r}+\frac{\mathrm{1}}{{r}}={u} \\ $$$${u}^{\mathrm{2}} +{u}−\left(\frac{\mathrm{211}}{{a}}+\mathrm{1}\right)=\mathrm{0} \\ $$
Commented by prakash jain last updated on 03/Sep/16
$$\mathrm{Since}\:\mathrm{all}\:\mathrm{terms}\:\mathrm{are}\:\mathrm{less}\:\mathrm{than}\:\mathrm{hundred} \\ $$$${r}<\mathrm{4}. \\ $$$$\mathrm{Only}\:\mathrm{option}\:\mathrm{for}\:{r}\:\mathrm{are}\:\mathrm{1},\mathrm{2},\mathrm{3} \\ $$$${no}\:{solution}\:{for}\:{sum}=\mathrm{211} \\ $$
Commented by 314159 last updated on 04/Sep/16
$${Thank}\:{a}\:{lot}! \\ $$