# Math Help - Geometric Series

1. ## Geometric Series

I have a problem with finding the sum of this geometric series. (the characters in italics are supposed to be smaller)

a1 = 2, a6 = 486, r = 3

So, applying the sum formula:

Sn = a1 (1 - r^n) / 1 - r

Sn = 2 (1 - 3^n) / 1 - r

I don't have n to figure it out with this formula, instead I have a6, the 6th term. What am I supposed to do with that?

2. Originally Posted by BlueStar
I have a problem with finding the sum of this geometric series. (the characters in italics are supposed to be smaller)

a1 = 2, a6 = 486, r = 3

So, applying the sum formula:

Sn = a1 (1 - r^n) / 1 - r

Sn = 2 (1 - 3^n) / 1 - r

I don't have n to figure it out with this formula, instead I have a6, the 6th term. What am I supposed to do with that?
what sum did they ask you to find? the sum of the first what terms? six? if so, n = 5, if you start your series counting from zero, which in this case, i think is what you did

the terms of a geometric series are given by: $\sum_{n = 1}^{ \infty} a_1 r^{n - 1}$ or $\sum_{n = 0}^{ \infty} a_1 r^{n}$

3. Hello, BlueStar!

Is that the original wording of the problem? . . . It is truly strange.
. . There is too much information.

If the first term is: $a_1 = 2$ and the common ratio is: $r = 3$
. . we can use: . $a_n = a_1r^{n-1}$ and find that: . $a_6 \:=\:2\!\cdot\!3^5 \:=\:486$

If the first term is: $a_1 = 2$ and the sixth term is: $a_6 = 486$
. . then: . $486 \:=\:2\!\cdot r^5\quad\Rightarrow\quad r = 3$

If the sixth term is: $a_6 = 486$ and the common ratio is: $r = 3$
. . then: . $486 \:=\:a_1\!\cdot\!3^5\quad\Rightarrow\quad a_1 = 2$

That is, with two of the facts, we can determine the third.
. . So why gives redundant statements?

On the other hand, they didn't tell us how many terms are in the series.
. . Did I say "strange"? .I meant silly.