if-the-sum-S-4-of-the-first-4-terms-of-a-G-P-is-24-and-the-sum-S-6-of-the-first-6-terms-is-30-find-the-first-term-and-common-ratio-

Question Number 34169 by Rio Mike last updated on 01/May/18

$$\:{if}\:{the}\:{sum}\:{S}_{\mathrm{4}} \:{of}\:{the}\:{first}\:\mathrm{4}\:{terms} \\ $$$${of}\:{a}\:{G}\:{P}\:{is}\:\mathrm{24}\:{and}\:{the}\:{sum}\:{S}_{\mathrm{6}} \:{of}\: \\ $$$${the}\:{first}\:\mathrm{6}\:{terms}\:{is}\:\mathrm{30}\:{find}\:{the}\: \\ $$$${first}\:{term}\:{and}\:{common}\:{ratio}.. \\ $$

Answered by MJS last updated on 02/May/18

...really a GP? You won′t like the solution... a_0 (1+r+r^2 +r^3 )=24 ⇒ a_0 =((24)/(1+r+r^2 +r^3 )) a_0 (1+r+r^2 +r^3 +r^4 +r^5 )=30 24((1+r+r^2 +r^3 +r^4 +r^5 )/(1+r+r^2 +r^3 ))=30 24((1+r^2 +r^4 )/(1+r^2 ))=30 r^4 −(1/4)r^2 −(1/4)=0 r^2 =(1/8)±((√(17))/8) r^2 >0 ⇒ r=±((√(1+(√(17))))/( (√8)))=−((√(2+2(√(17))))/4) ∨ ((√(2+2(√(17))))/4) a_0 =(3/8)(4+(√(2+2(√(17)))))(23+(√(17))) ∨ (3/8)(4−(√(2+2(√(17)))))(23+(√(17))) (r≈−.800243∧a_0 ≈73.2423)∨ (r≈.800243∧a_0 ≈8.12706)

$$…\mathrm{really}\:\mathrm{a}\:\mathrm{GP}?\:\mathrm{You}\:\mathrm{won}'\mathrm{t}\:\mathrm{like}\:\mathrm{the}\:\mathrm{solution}… \\ $$$${a}_{\mathrm{0}} \left(\mathrm{1}+{r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} \right)=\mathrm{24}\:\Rightarrow\:{a}_{\mathrm{0}} =\frac{\mathrm{24}}{\mathrm{1}+{r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} } \\ $$$${a}_{\mathrm{0}} \left(\mathrm{1}+{r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} +{r}^{\mathrm{4}} +{r}^{\mathrm{5}} \right)=\mathrm{30} \\ $$$$\mathrm{24}\frac{\mathrm{1}+{r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} +{r}^{\mathrm{4}} +{r}^{\mathrm{5}} }{\mathrm{1}+{r}+{r}^{\mathrm{2}} +{r}^{\mathrm{3}} }=\mathrm{30} \\ $$$$\mathrm{24}\frac{\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} }{\mathrm{1}+{r}^{\mathrm{2}} }=\mathrm{30} \\ $$$${r}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{4}}{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}}\pm\frac{\sqrt{\mathrm{17}}}{\mathrm{8}} \\ $$$${r}^{\mathrm{2}} >\mathrm{0}\:\Rightarrow\:{r}=\pm\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{17}}}}{\:\sqrt{\mathrm{8}}}=−\frac{\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{17}}}}{\mathrm{4}}\:\vee\:\frac{\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{17}}}}{\mathrm{4}} \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{4}+\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{17}}}\right)\left(\mathrm{23}+\sqrt{\mathrm{17}}\right)\:\vee\:\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{4}−\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{17}}}\right)\left(\mathrm{23}+\sqrt{\mathrm{17}}\right) \\ $$$$\left({r}\approx−.\mathrm{800243}\wedge{a}_{\mathrm{0}} \approx\mathrm{73}.\mathrm{2423}\right)\vee \\ $$$$\left({r}\approx.\mathrm{800243}\wedge{a}_{\mathrm{0}} \approx\mathrm{8}.\mathrm{12706}\right) \\ $$

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