Progressions

**mastermin346** · December 23rd 2009, 07:22 AM

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.

**mathaddict** · December 23rd 2009, 07:27 AM

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

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

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

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

Solve for r then .

**e^(i*pi)** · December 23rd 2009, 08:08 AM

Originally Posted by mathaddict

HI

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

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

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

Solve for r then .

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

$(a-b)^3 = (a-b)(a^2+ab+b^2)$

**Soroban** · December 23rd 2009, 08:38 AM

Hello, 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.

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

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

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

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

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