# Math Help - Geometric Series issue.

1. ## Geometric Series issue.

Having a bit of trouble with the following problem:

The sum of the first three terms of a geometric series is 21. The sum of terms 4 through 6 is 168. Find the first three terms.

Would greatly appreciate help.

2. ## Re: Geometric Series issue.

Hello, Ventus!

Do you know anything about Geometric Series?

The sum of the first three terms of a geometric series is 21.
The sum of terms 4 through 6 is 168.
Find the first three terms.

The first six terms are: . $a,\; ar,\; ar^2,\; ar^3,\; ar^4,\; ar^5$

We are told: . $\begin{Bmatrix}a + ar + ar^2 \:=\: 21 & \Rightarrow & a(1+r + r^2) \:=\: 21 & [1] \\ \\[-3mm] ar^3 + ar^4 + ar^5 \:=\: 168 & \Rightarrow & ar^3(1+r+r^2) \:=\: 168 & [2] \end{Bmatrix}$

Divide [2] by [1]: . $\frac{ar^3(1+r+r^2)}{a(1+r+r^2)} \:=\:\frac{168}{21}$

. . And we have: . $r^3 \:=\:8 \quad\Rightarrow\quad \boxed{r \:=\:2}$

Substitute into [1]: . $a(1+2+2^2) \:=\:21 \quad\Rightarrow\quad 7a \:=\:21 \quad\Rightarrow\quad \boxed{a \:=\:3}$

Therefore, the first three terms are: . $(a,\:ar,\:ar^2) \;=\;(3,\:6,\:12)$

3. ## Re: Geometric Series issue.

Originally Posted by Soroban
Hello, Ventus!

Do you know anything about Geometric Series?

The first six terms are: . $a,\; ar,\; ar^2,\; ar^3,\; ar^4,\; ar^5$

We are told: . $\begin{Bmatrix}a + ar + ar^2 \:=\: 21 & \Rightarrow & a(1+r + r^2) \:=\: 21 & [1] \\ \\[-3mm] ar^3 + ar^4 + ar^5 \:=\: 168 & \Rightarrow & ar^3(1+r+r^2) \:=\: 168 & [2] \end{Bmatrix}$

Divide [2] by [1]: . $\frac{ar^3(1+r+r^2)}{a(1+r+r^2)} \:=\:\frac{168}{21}$

. . And we have: . $r^3 \:=\:8 \quad\Rightarrow\quad \boxed{r \:=\:2}$

Substitute into [1]: . $a(1+2+2^2) \:=\:21 \quad\Rightarrow\quad 7a \:=\:21 \quad\Rightarrow\quad \boxed{a \:=\:3}$

Therefore, the first three terms are: . $(a,\:ar,\:ar^2) \;=\;(3,\:6,\:12)$

Oh I see, I neglected to factor out a and ar^3.

Thank you very much! I'll be going into my test with much added confidence.