1. ## geometric sequence

find the sum of the first 12 terms of the geometric sequence 2,6,12,24

2. Hello, Shorty!

I assume there's a typo in the problem.
. . As given, it is not a geometric sequence.

You're expected to know these formulas for geometric sequences.

The $n^{th}$ term: . $a_n\:=\:ar^{n-1}$

The sum of the first $n$ terms: . $S_n\:=\:a\frac{r^n - 1}{r - 1}$

. . where $a$ is the first term and $r$ is the common ratio.

Find the sum of the first 12 terms of the geometric sequence: .3, 6, 12, 24 . . .

The first term is $a = 3.$
The common ratio is $r = 2.$

Then: . $S_{12} \;= \;3\,\frac{2^{12} - 1}{2 - 1} \:=\:3(4095) \;= \;\boxed{12,285}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The very least you could do is crank out the first 12 terms and add!

. . $3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 + 1536 + 3072 + 6144$

What was stopping you?

3. Originally Posted by SHORTY
find the sum of the first 12 terms of the geometric sequence 2,6,12,24
There's a problem with your sequence: the ratio of the terms aren't the same. For example 6/2 = 3, but 12/6 = 2. These ratios must be the same in any geometric sequence.

-Dan