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Question Number 65901 by byaw last updated on 05/Aug/19

Answered by MJS last updated on 05/Aug/19

a_0   a_3 =a_0 +2α  a_8 =a_0 +7α  g_0 =a_0   g_1 =γg_0 =a_3   g_2 =γg_1 =a_8   γ=(a_3 /a_0 )=(a_8 /a_3 )  ⇒ a_3 ^2 =a_0 a_8   (a_0 +2α)^2 =a_0 (a_0 +7α)  4a_0 +4α=7a_0   α=(3/4)a_0   g_0 =a_0   g_1 =(5/2)a_0   g_2 =((25)/4)a_0   ⇒ γ=(5/2)  g_0 +g_1 +g_2 =((39)/4)a_0 =28 ⇒ a_0 =((112)/(39))=g_0   ⇒ α=((28)/(13))

$${a}_{\mathrm{0}} \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{0}} +\mathrm{2}\alpha \\ $$$${a}_{\mathrm{8}} ={a}_{\mathrm{0}} +\mathrm{7}\alpha \\ $$$${g}_{\mathrm{0}} ={a}_{\mathrm{0}} \\ $$$${g}_{\mathrm{1}} =\gamma{g}_{\mathrm{0}} ={a}_{\mathrm{3}} \\ $$$${g}_{\mathrm{2}} =\gamma{g}_{\mathrm{1}} ={a}_{\mathrm{8}} \\ $$$$\gamma=\frac{{a}_{\mathrm{3}} }{{a}_{\mathrm{0}} }=\frac{{a}_{\mathrm{8}} }{{a}_{\mathrm{3}} } \\ $$$$\Rightarrow\:{a}_{\mathrm{3}} ^{\mathrm{2}} ={a}_{\mathrm{0}} {a}_{\mathrm{8}} \\ $$$$\left({a}_{\mathrm{0}} +\mathrm{2}\alpha\right)^{\mathrm{2}} ={a}_{\mathrm{0}} \left({a}_{\mathrm{0}} +\mathrm{7}\alpha\right) \\ $$$$\mathrm{4}{a}_{\mathrm{0}} +\mathrm{4}\alpha=\mathrm{7}{a}_{\mathrm{0}} \\ $$$$\alpha=\frac{\mathrm{3}}{\mathrm{4}}{a}_{\mathrm{0}} \\ $$$${g}_{\mathrm{0}} ={a}_{\mathrm{0}} \\ $$$${g}_{\mathrm{1}} =\frac{\mathrm{5}}{\mathrm{2}}{a}_{\mathrm{0}} \\ $$$${g}_{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{4}}{a}_{\mathrm{0}} \\ $$$$\Rightarrow\:\gamma=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${g}_{\mathrm{0}} +{g}_{\mathrm{1}} +{g}_{\mathrm{2}} =\frac{\mathrm{39}}{\mathrm{4}}{a}_{\mathrm{0}} =\mathrm{28}\:\Rightarrow\:{a}_{\mathrm{0}} =\frac{\mathrm{112}}{\mathrm{39}}={g}_{\mathrm{0}} \\ $$$$\Rightarrow\:\alpha=\frac{\mathrm{28}}{\mathrm{13}} \\ $$

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