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July 6th, 2012, 04:52 AM   #1
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Arithmetic Progression

Can anyone solve this, I have the exam in about 2 and half hours... so please if any one know how to solve it do it ASAP

Translated request: "Find arithmetic progression, if:"
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 July 6th, 2012, 06:19 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: [A.S.A.P] Arithmetic Progression The hardest part here is keeping the algebra under control. Let $c=x_3, \;d=|x_{k+1}-x_k|$. Then $(c-d)(c+d)=2, \;(c-2d)(c+3d)=\frac32$ Expanding, $c^2=d^2+2,\; c^2+cd-6d^2=\frac32$ Substituting the first into the second, $d\sqrt{d^2+2}=-\frac12+5d^2$ Notice that I've put the square root by itself on one side, so as to simplify the next step. Square both sides and rearrange to get $24d^4-7d^2+\frac14=0$ Solving the quadratic in d², you get $d=\frac{1}{2\sqrt6}\text{ or }\frac12$. Hence $c=\frac{-7}{2\sqrt6}\text{ or }\frac32$ and all the terms can now be found. Two more solutions can be made by flipping the signs of c and d simultaneously.
 July 6th, 2012, 06:29 AM #3 Senior Member   Joined: Aug 2008 Posts: 113 Thanks: 0 Re: [A.S.A.P] Arithmetic Progression $x_1= a$ $x_n= a + d(n-1)$ $x_6= a + 5d$ $x_2= a + d$ $x_4= a + 3d$ Given: $a(a+5d)= \frac{3}{2}$ Or, $a^2 + 5ad= \frac{3}{2}$ Or, $d= \frac {\frac{3}{2} - a^2}{5a}$ $(a+d)(a+3d)= 2$ Or, $a^2 + 4ad + 3d^2= 2$ Or, $a^2 + 4a \frac {\frac{3}{2} - a^2}{5a} +3 \left(\frac {\frac{3}{2} - a^2}{5a}\right)^2= 2$ Or, $25a^4 + 20a^2 {(\frac{3}{2} - a^2)} + 3 (\frac{3}{2} - a^2)^2= 2$ Or, $25a^4 + 30a^2 - 20a^4 + 3(\frac{9}{4} + a^4 - 3a^2)= 50a^2$ Or, $8a^4 - 29a^2 + \frac{27}{4}= 0$ Or, $8z^2 - 29z + \frac{27}{4}= 0$ where $z= a^2$ Or, $z= \frac{29 \pm \sqrt{29^2 - 4 \cdot 8 \cdot \frac {27}{4}}}{2 \cdot 8} = \frac{29 \pm 25}{16} = \frac{1}{4}, \frac{27}{8}$ Can you solve the rest now? Best regards,
 July 6th, 2012, 06:38 AM #4 Senior Member   Joined: Aug 2008 Posts: 113 Thanks: 0 Re: Arithmetic Progression I think there are 4 solutions. Please verify if all of them are valid. The ordered pairs of values of $(a,d)$ are: $\left(\frac{1}{2},\;\frac{1}{2}\right)$ $\left(-\frac{1}{2},\;-\frac{1}{2}\right)$ $\left(\frac{3\sqrt{6}}{4},\;-\frac{\sqrt{6}}{12}\right)$ $\left(-\frac{3\sqrt{6}}{4},\;\frac{\sqrt{6}}{12}\right)$ Best regards,

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