The sum of the firsts two terms of a geometric progression..?

• Feb 6th 2011, 12:34 PM
manexa
The sum of the firsts two terms of a geometric progression..?
I hope this is the right section. If not please move.

The sum of the firsts two terms of a geometric progression is six times the magnitude of the third term:

i.e. S2 = 6*T3

This is only possible for two values of the common ratio, r = 0.5 is one value.
Calculate the other value correct to three decimal places.

Any help with this would be appreciated. If you could give a step by step explanation on how its done it would be great.
• Feb 6th 2011, 12:59 PM
$\displaystyle T_2=rT_1$

$\displaystyle T_3=r^2T_1$

$\displaystyle S_2=T_1+rT_1$

$\displaystyle 6r^2T_1=rT_1+T_1\Rightarrow\ T_1\left(6r^2-r-1)=0$

$\displaystyle 6r^2-r-1=0\Rightarrow\ (3r+1)(2r-1)=0$
• Feb 6th 2011, 06:49 PM
Wilmer
90 *(-1/3) -30 *(-1/3) 10
90 + (-30) = 60
10 * 6 = 60
• Feb 8th 2011, 02:28 PM
manexa
Oh I see that it is a quadratic equation. :)

I like to use the b+- formula.

Thanks for the help.