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Question Number 7541 by 314159 last updated on 03/Sep/16

Commented by prakash jain last updated on 03/Sep/16

$$a(1+r+r^2+r^3+r^4)=211$$$$aandrintegers$$$$211=1\times 211\Rightarrow a=1$$$$\frac{r^5-1}{r-1}=211$$

Commented by prakash jain last updated on 03/Sep/16

$$211\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{any}\mathrm{valid}\mathrm{answer}.$$$$\mathrm{I}\mathrm{think}\mathrm{question}\mathrm{should}\mathrm{have}\mathrm{been}$$$$\mathrm{sum}=121$$$$1+3+9+27+81=121$$

Commented by prakash jain last updated on 03/Sep/16

$$\mathrm{Let}\mathrm{the}\mathrm{GP}\mathrm{be}$$$$\frac{a}{r^2},\frac{a}{r},a,ar,{ar}^2$$$$\frac{a}{r^2}+\frac{a}{r}+a+ar+{ar}^2=211$$$$r^2+\frac{1}{r^2}+r+\frac{1}{r}+1=\frac{211}{a}$$$${(r+\frac{1}{r})}^2+(r+\frac{1}{r})-1=\frac{211}{a}$$$$r+\frac{1}{r}=u$$$$u^2+u-(\frac{211}{a}+1)=0$$

Commented by prakash jain last updated on 03/Sep/16

$$\mathrm{Since}\mathrm{all}\mathrm{terms}\mathrm{are}\mathrm{less}\mathrm{than}\mathrm{hundred}$$$$r<4.$$$$\mathrm{Only}\mathrm{option}\mathrm{for}r\mathrm{are}1,2,3$$$$nosolutionforsum=211$$

Commented by 314159 last updated on 04/Sep/16

$$Thankalot!$$

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