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Question Number 7855 by 314159 last updated on 21/Sep/16

Commented by 123456 last updated on 21/Sep/16

$$a_1,a_2,a_3\Rightarrow a_3-a_2=a_2-a_1,a_1\ne a_2$$$$a_1,a_2,b_3,a_3\Rightarrow \frac{a_3}{b_3}=\frac{b_3}{a_2}=\frac{a_2}{a_1}$$$$\mathrm{\Sigma }b=3+\sqrt{5}$$$$\frac{a_1}{1-\frac{a_2}{a_1}}=3+\sqrt{5}\mid \frac{a_2}{a_1}\mid <1$$

Commented by Rasheed Soomro last updated on 21/Sep/16

$$a,ar,{ar}^2,{ar}^3,...[a:firstterm,r:commonratio]$$$$a,ar,{ar}^{3}areinAP,arisAMbetweenaand{ar}^3$$$$\therefore ar=\frac{a+{ar}^3}{2}\Rightarrow 2r=1+r^3\Rightarrow r^3-2r+1=0$$$$S=\frac{a}{1-r}=3+\sqrt{5}[S=a+ar+{ar}^2+...]..........(\ast )$$$$Anattempttofactrize{r}^3-2r+1$$$$r^3-2r+1=0$$$$2r^3-2r+1-r^3=0$$$$2r(r^2-1)-(r^3-1)=0$$$$2r(r-1)(r+1)-(r-1)(r^2+r+1)=0$$$$(r-1)[2r(r+1)-(r^2+r+1)]=0$$$$r-1=0\mid 2r^2+2r-r^2-r-1=0$$$$r=1\mid r^2+r-1=0$$$$r=1isdiscardablebecauseSisinfinitythere.$$$$So{r}^2+r-1=0[r\ne 1]$$$$r=\frac{-1\pm \sqrt{1+4}}{2}=\frac{1\pm \sqrt{5}}{2}$$$$Asr<1Soas\frac{1+\sqrt{5}}{2}>0,Hencer=\frac{1-\sqrt{5}}{2}istheonlyvalue.$$$$From(\ast )$$$$\frac{a}{1-r}=3+\sqrt{5}$$$$a=(3+\sqrt{5})(1-r)$$$$a=(3+\sqrt{5})(1-\frac{1-\sqrt{5}}{2})$$$$a=(3+\sqrt{5})\left(\frac{2-1+\sqrt{5}}{2}\right)=(3+\sqrt{5})\left(\frac{1+\sqrt{5}}{2}\right)$$$$a=\frac{(3+\sqrt{5})(1+\sqrt{5})}{2}$$$$a=\frac{3+3\sqrt{5}+\sqrt{5}+5}{2}$$$$=4+2\sqrt{5}$$$$Sothefirsttermis4+2\sqrt{5}$$$$$$

Commented by 314159 last updated on 22/Sep/16

$$\mathbf{Thank}\mathbf{a}\mathbf{lot}!$$

Commented by lepan last updated on 06/Oct/16

$$a=-1-\sqrt{5}$$

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