1. ## Geometric sequence

Hello all!

I have a problem: whenever I am doing word problems involving geometric or arithmetic sequences or series, I always end up in a pair of simultaneous equations that I do not know how to solve. For instance, I am currently doing the following problem

Problem. The difference of the fifth term and the second term of a geometric series is equal to 156, and the the difference of the seventh term and the fourth term is equal to 1404. Find the first term of the geometric series and the common ratio.

Ok, so I start by writing a pair of equations.

Step 1.

$\displaystyle u_5-u_2=156\ (1)$ and $\displaystyle u_7-u_4=1404\ (2)$,
and then I plug in the standard definition of the [tex]nth[tex] terms of a geometric series $\displaystyle u_n=aq^{n-1}$, where $\displaystyle a$ is the first term and $\displaystyle q$ is the common ratio.

I get the following set of equations

$\displaystyle aq^4-aq=156$ and $\displaystyle aq^6-aq^2=1404$, but I do not know any technique to solve such a pair of simultaneous equations. I tried adding them together, solving for $\displaystyle a$, and other things, but I always end up in really complicated equations that I most certainly am not able to solve.

Step 2.
I notice that in both cases the difference is equal to two terms.
In the case of eq. (1) the difference is equal to $\displaystyle u_3+u_4=156$, and in the case of eq. (2) the difference is equal to $\displaystyle u_5+u_6=1404$.
So then I let $\displaystyle u_3=b=aq^2$ and I get the following equations

$\displaystyle \frac{b(1-q^2)}{1-q}=25$ and $\displaystyle \frac{bq^2(1-q^2)}{1-q}=1404$
So then I divide one by the other to get
$\displaystyle q^2=9$ and $\displaystyle q=\pm 3$

Step 3.
$\displaystyle b+bq=156$$\displaystyle 4b=156$
$\displaystyle b=39$

since $\displaystyle b=aq^2$
$\displaystyle a=\frac{13}{3}$
But according to my book this answer is wrong.
There is something wrong with my method, right? Is there a better way to do this problem?
Thanks to anyone who replies.

2. $\displaystyle \left\{\begin{array}{ll}aq^4-aq=156\\aq^6-1q^3=1404\end{array}\right.\Rightarrow\left\{\begi n{array}{ll}aq(q^3-1)=156\\aq^3(q^3-1)=1404\end{array}\right.$

Then $\displaystyle \frac{aq(q^3-1)}{aq^3(q^3-1)}=\frac{156}{1404}\Rightarrow\frac{1}{q^2}=\frac {1}{9}\Rightarrow q=\pm 3$

Can you continue?