Find the numerical value of the first 3 terms of a geometric sequence

Question

The first 3 terms in a geometric sequence are $x, x+5, x+9$.

With this information, determine the numerical value of these terms.

We didn't go over this in class. I'm not sure what formula to use.

$a=x$

$r=?$

Any help is greatly appreciated.

Answer 1

An arithmetic sequence is $x,\ x+k,\ x+2k,\ x+3k,\ \cdots\quad$ where $k$ is called the increment.

A geometric sequence is $x,\ xr,\ xr^2\ xr^3,\ \cdots\quad$ where $r$ is called the reason.

Here you have to solve $\begin{cases} x+5 = xr\\x+9 = xr^2\end{cases}$

So $x+9=xr^2=(xr)r=(x+5)r=xr+5r=x+5+5r\iff 5r=4\iff r=\frac 45$

And $x+5=\frac 45x\iff 25=4x-5x=-x\iff x=-25$

Verification : $\begin{cases}x+5=-25+5=-20=-25\times\frac 45\\x+9=-25+9=-16=-25\times\frac{16}{25}\end{cases}$

Answer 2

We start off with the fact that

$$r=\dfrac{x+5}x=\dfrac{x+9}{x+5}$$

$$\begin{align}(x+5)^2&=x(x+9)\\x^2+10x+25&=x^2+9x\\10x+25&=9x\\x&=-25\end{align}$$

Now find $(x+5),(x+9)$

Answer 3

The term of geometric sequence follow the pattern $$ x, rx, r^2x,... $$In our situation we have $$ x, x+5, x+9,...$$ Note that and $$x+5=rx$$ and $$ x+9 = r(x+5).$$ Therefore, $$ \frac {x+5}{x}=\frac{x+9}{x+5}.$$Cross multiplication and solving for $x$ results in $x=-25.$ Therefor our terms are $$ -25,-20,-16,...$$ The value of $r$ is found to be $r=(-20)/(-25)=4/5$