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!
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.