Geometric Series with only one known term

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B Ann

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A geometric series has a second term of 6. The sum of its fist 3 terms is -14. Find the fourth term.

I started with the Sn formula: Sn=(x(r^n-1))/(r-1). Since the first term (x) = 6/r,

(1) -14= 6(r^3-1)/(r(r-1))
(2) -14= (6r^3-6)/(r^2-r)
(3) -14=(6r^2-6)/-r
(4) 6r^2-14r-6 .... the quadratic equation gave me r= 0.37 or -2.7
(5) The fourth number should be equal to 6r, meaning either 2.22 or -16.23

I know my math is wrong, and I think my mistake is in step (3). I would appreciate if someone could correct me.
 
B Ann said:
A geometric series has a second term of 6. The sum of its fist 3 terms is -14. Find the fourth term.

I started with the Sn formula: Sn=(x(r^n-1))/(r-1). Since the first term (x) = 6/r,

(1) -14= 6(r^3-1)/(r(r-1))
(2) -14= (6r^3-6)/(r^2-r)
(3) -14=(6r^2-6)/-r
(4) 6r^2-14r-6 .... the quadratic equation gave me r= 0.37 or -2.7
(5) The fourth number should be equal to 6r, meaning either 2.22 or -16.23

I know my math is wrong, and I think my mistake is in step (3). I would appreciate if someone could correct me.
Click to expand...

Instead of using sum equation - let's do the following:

first 3 terms: 6/r , 6 & 6r .... the fourth term is 6r^2

6/r + 6 + 6r = -14

6/r + 6r = -20

6r^2 + 20r + 6 = 0 ..........use quadratic equation

r = -10/6 +/- [sqrt{400-144}]/12 = [-5/3] +/- [4/3] = -1/3 or -3

continue.....
 
Thank you,
and to help whoever ends up interested by this thread in the future,
6r^2= 0.666... or 54.
 
There are two solutions (corresponding to two possible values of r)

-2, 6, -18, 54...

or

-18, 6, -2, 2/3.....
 
B Ann said:
A geometric series has a second term of 6. The sum of its fist 3 terms is -14. Find the fourth term.
Click to expand...
Look you have \(\displaystyle a+ar+ar^2=-14\) and \(\displaystyle ar=6\) so that \(\displaystyle a=\frac{6}{r}\).
From which we have \(\displaystyle \frac{6}{r}+\left(\frac{6}{r}\right)r+\left(\frac{6}{r}\right)r^2=-14\)
Can you work with that?
 
pka said:
Look you have \(\displaystyle a+ar+ar^2=-14\) and \(\displaystyle ar=6\) so that \(\displaystyle a=\frac{6}{r}\).
From which we have \(\displaystyle \frac{6}{r}+\left(\frac{6}{r}\right)r+\left(\frac{6}{r}\right)r^2=-14\)
Can you work with that?
Click to expand...

Why is the difference between each number multiplicative? I thought it was "a + (a+d) + (a + 2d) + (a + 3d) +... "
 
Alright, I just looked it up and I see the rule of it having to be a multiplicative difference to be considered a geometric sequence. Nevermind to ALL of my comments. Sorry, I'll search before I post all these things next time.
 
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