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On-a-path-there-was-an-odd-number-of-rocks-with-a-distance-of-10m-between-them-we-want-to-put-them-all-where-the-middle-one-is-if-we-start-to-collect-from-one-of-the-ends-and-when-we-finish-we-

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Question Number 182704 by HeferH last updated on 13/Dec/22
On a path there was an odd number of rocks with a distance of 10m between them, we want to put them all where the middle one is, if we start to collect from one of the ends and when we finish we have walked 3km in total, how many rocks are there?
$${On}\:{a}\:{path}\:{there}\:{was}\:{an}\:{odd}\:{number}\:{of}\:{rocks} \\ $$$$\:{with}\:{a}\:{distance}\:{of}\:\mathrm{10}{m}\:{between}\:{them},\:{we}\: \\ $$$$\:{want}\:{to}\:{put}\:{them}\:{all}\:{where}\:{the}\:{middle}\:{one}\:{is}, \\ $$$$\:{if}\:{we}\:{start}\:{to}\:{collect}\:{from}\:{one}\:{of}\:{the}\:{ends}\:{and} \\ $$$$\:{when}\:{we}\:{finish}\:{we}\:{have}\:{walked}\:\mathrm{3}{km}\:{in}\:{total}, \\ $$$$\:{how}\:{many}\:{rocks}\:{are}\:{there}? \\ $$
Answered by TheSupreme last updated on 13/Dec/22
if n=1 S_1 =0 if n=3 S_3 =3∗10 if n=5 S_5 =S_3 +10+2∗3∗10 S_(2n+1) =S_(2n−1) +10∗(n−1)+3∗n∗10 S_(2n+1) =Σ_(i=1) ^n 10(i−1)+30∗i=Σ_(i=1) ^n 40i−10 S_(2n+1) =40 ((n(n+1))/2)−10n 3000=20n^2 +10n=10n(2n+1) n=12
$${if}\:{n}=\mathrm{1}\:{S}_{\mathrm{1}} =\mathrm{0} \\ $$$${if}\:{n}=\mathrm{3}\:{S}_{\mathrm{3}} =\mathrm{3}\ast\mathrm{10} \\ $$$${if}\:{n}=\mathrm{5}\:{S}_{\mathrm{5}} ={S}_{\mathrm{3}} +\mathrm{10}+\mathrm{2}\ast\mathrm{3}\ast\mathrm{10} \\ $$$${S}_{\mathrm{2}{n}+\mathrm{1}} ={S}_{\mathrm{2}{n}−\mathrm{1}} +\mathrm{10}\ast\left({n}−\mathrm{1}\right)+\mathrm{3}\ast{n}\ast\mathrm{10} \\ $$$${S}_{\mathrm{2}{n}+\mathrm{1}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{10}\left({i}−\mathrm{1}\right)+\mathrm{30}\ast{i}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{40}{i}−\mathrm{10} \\ $$$${S}_{\mathrm{2}{n}+\mathrm{1}} =\mathrm{40}\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\mathrm{10}{n} \\ $$$$\mathrm{3000}=\mathrm{20}{n}^{\mathrm{2}} +\mathrm{10}{n}=\mathrm{10}{n}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$${n}=\mathrm{12} \\ $$$$ \\ $$
Answered by mr W last updated on 13/Dec/22
say there are 2n+1 rocks with distance d between them. total distance to walk: L=4(d+2d+3d+...+nd)−nd=n(2n+1)d 3000=n(2n+1)10 2n^2 +n−300=0 ⇒n=12 i.e. there are 2×12+1=25 rocks
$${say}\:{there}\:{are}\:\mathrm{2}{n}+\mathrm{1}\:{rocks}\:{with}\:{distance}\:{d} \\ $$$${between}\:{them}. \\ $$$${total}\:{distance}\:{to}\:{walk}: \\ $$$${L}=\mathrm{4}\left({d}+\mathrm{2}{d}+\mathrm{3}{d}+…+{nd}\right)−{nd}={n}\left(\mathrm{2}{n}+\mathrm{1}\right){d} \\ $$$$\mathrm{3000}={n}\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{10} \\ $$$$\mathrm{2}{n}^{\mathrm{2}} +{n}−\mathrm{300}=\mathrm{0} \\ $$$$\Rightarrow{n}=\mathrm{12} \\ $$$${i}.{e}.\:{there}\:{are}\:\mathrm{2}×\mathrm{12}+\mathrm{1}=\mathrm{25}\:{rocks} \\ $$

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