# Progressions

• Dec 23rd 2009, 07:22 AM
mastermin346
Progressions
The sum of the first 3 terms of a geometric progression is 13 times its first term.Find the possible values of the common ratio of the geometric progression.
• Dec 23rd 2009, 07:27 AM
Quote:

Originally Posted by mastermin346
The sum of the first 3 terms of a geometric progression is 13 times its first term.Find the possible values of the common ratio of the geometric progression.

HI

$\displaystyle S_3=13a$ , where a is the first term

$\displaystyle \frac{a(1-r^3)}{1-r}=13a$

Don worry bout the a because they will eventually cancel off.

Solve for r then .
• Dec 23rd 2009, 08:08 AM
e^(i*pi)
Quote:

HI

$\displaystyle S_3=13a$ , where a is the first term

$\displaystyle \frac{a(1-r^3)}{1-r}=13a$

Don worry bout the a because they will eventually cancel off.

Solve for r then .

You should also be able to cancel $\displaystyle a$ immediately and then use the difference of two cubes to cancel $\displaystyle (1-r)$ given that $\displaystyle r \neq 1$

$\displaystyle (a-b)^3 = (a-b)(a^2+ab+b^2)$
• Dec 23rd 2009, 08:38 AM
Soroban
Hello, mastermin346!

Quote:

The sum of the first 3 terms of a geometric progression is 13 times its first term.
Find the possible values of the common ratio of the geometric progression.

The first three terms are: .$\displaystyle a,\:ar,\:ar^2$

Their sum is 13 times the first term: .$\displaystyle a + ar + ar^2 \:=\:13a$

We have: .$\displaystyle r^2 + r - 12 \:=\:0 \quad\Rightarrow\quad (r - 3)(r + 4) \:=\:0$

Therefore: .$\displaystyle r \;=\;3\text{ or }-4$