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The fourth term of a G.P is $3$ and the sixth term is $147$. Find the first $3$ terms of the two possible geometric progressions.

Can you help me find $a$ and $r$? It is too complicated. I took two simultaneous equations

$$ar^3=3 ------- (1)$$

$$ar^5=147------- (2)$$

$$r= \sqrt[3]{\frac{3}{a}}------(3)$$

I did this, and then it gets complicated because of that cubic root...

i got $a=0.672$ and $r=1.67$ which is wrong. Help!

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Divide $ar^5$ by $ar^3$. You should get $r^2=\frac{147}{3}$. Thus $r=\pm 7$. Now for each value of $r$ the appropriate $a$ is not hard to find.

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  • $\begingroup$ that means a is right? $\endgroup$
    – Kiara
    Commented Jul 11, 2014 at 14:27
  • $\begingroup$ should i divide (2) / (1)? $\endgroup$
    – Kiara
    Commented Jul 11, 2014 at 14:29
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    $\begingroup$ There are two cases. If $r=7$, then $a=\frac{3}{r^3}=\frac{3}{343}$. The other case gives you $a=-\frac{3}{343}$. But first we needed $r$, which is obtained by finding $\frac{ar^5}{ar^3}$. Alternately we could find $a$ first, but that is somewhat more messy. $\endgroup$
    – André Nicolas
    Commented Jul 11, 2014 at 14:30
  • $\begingroup$ alright...lemme try :) thanks! $\endgroup$
    – Kiara
    Commented Jul 11, 2014 at 14:31
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    $\begingroup$ You are welcome. The answer I gave should be enough for you to finish the rest of the calculation. $\endgroup$
    – André Nicolas
    Commented Jul 11, 2014 at 14:33

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