A spinner has four sectors whose angles form a G.P . The smallest angle is \displaystyle {{24}^{\circ }} and the sector are numbered as 1,2,3,4 in ascending order of their degree measures . If the arrow is spun 720 times what final score would you expect when all the individual scores are added together ?
Solution
Let the four angles be \displaystyle {{24}^{\circ }},{{24}^{\circ }}r,{{24}^{\circ }}{{r}^{2}},{{24}^{\circ }}{{r}^{3}} .
\displaystyle {{24}^{\circ }}+{{24}^{\circ }}r+{{24}^{\circ }}{{r}^{2}}+{{24}^{\circ }}{{r}^{3}}={{360}^{\circ }}
\displaystyle {{1}^{\circ }}+r+{{r}^{2}}+{{r}^{3}}={{15}^{\circ }}
\displaystyle {{r}^{3}}+{{r}^{2}}+r-{{14}^{\circ }}=0
\displaystyle (r-{{2}^{\circ }})({{r}^{2}}+3r+{{7}^{\circ }})=0
\displaystyle r-{{2}^{\circ }}=0\ \ \ \ (or)\ \ \ \ {{r}^{2}}+3r+{{7}^{\circ }}=0
\displaystyle \ \ \ \ \ \ r={{2}^{\circ }}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (reject)\ (\because {{b}^{2}}-4ac<0)
\displaystyle \text{The four angles are 2}{{\text{4}}^{\circ }},{{48}^{\circ }},{{96}^{\circ }}\text{ and }{{192}^{\circ }}.
\displaystyle \text{Final score = }(1\times \frac{1}{{15}}\times 720)+(2\times \frac{2}{{15}}\times 720)+
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3\times \frac{4}{{15}}\times 720)+(4\times \frac{8}{{15}}\times 720)
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =720(\frac{1}{{15}}+\frac{4}{{15}}+\frac{{12}}{{15}}+\frac{{32}}{{15}})
\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =720\times \frac{{49}}{{15}}=2352
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