Geometric Progression with just difference and not ratio Its given that "In a certain geometric series, the fourth term exceeds the third term by $2$ and exceeds the second term by $5$" and I am supposed to find the third term of this geometric progression.
So lets say I let $g_1$ be the first term, $g_2$ be the second term, $g_3$ be the third term and $g_4$ be the fourth term.
I know I will get these 2 eqns
$$g_4-g_3 = 2$$
$$g_4-g_2 = 5$$
2 eqns, 3 unknowns, cannot be solved.
However, in a geometric sequence, they are related by common ratio $r$. How would then such a question be solved?
Thanks!
 A: Note that $g_2 = g_1r$ where $r$ is the common ratio. Likewise, $g_3 = g_2r$, but as $g_2 = g_1r$ we have $g_3 = g_1r^2$. In general, $g_{n+1} = g_1r^n$. Using this expression for $g_3$, $g_4$, and $g_5$, you will end up with two equations with two unknowns, $g_1$ and $r$.
Alternatively, you can use the same idea to write both $g_4$ and $g_2$ in terms of your desired quantity $g_3$ and the common ratio $r$. That is, use the fact that $g_4 = rg_3$ and $g_2 = \dfrac{g_3}{r}$. Then when you solve the two equations, you will get $r$ and $g_3$. In the previous method, you'd find $r$ and $g_1$, then have to calculate $g_3 = g_1r^2$.
A: You can express all of the $g_n$ using the common ratio:
$$g_n=ar^n$$
or you can express just $g_4$ in terms of $g_2$ and $g_3$:
$$g_4=\frac{g_3^2}{g_2}$$
Either way, substitute the formula into the equations you already have, and solve for the remaining two unknowns.
A: One can do it without even writing down equations:
The increase from $g_2$ to $g_3$ is $3$ (namely $5-2$).
The increase from $g_3$ to $g_4$ is $2$.
Because the progression is geometric, these increases are proportional to $g_2$ and $g_3$, respectively. So $g_3$ must be two thirds of $g_2$.
Which number increases by $3$ when you take a third of it away? The only possibility is that $g_2$ must be $-9$, and then $g_3$ is $-6$.
A: $g_4 - g_2 = 5$, so $r^2 g_2 - g_2 = 5$.
$g_4 - g_3 = 2$, so $r^2 g_2-rg_2=2$.
$$5= r^2g_2 - g_2 = (r^2-1)g_2=(r-1)(r+1)g_2.\tag1$$
$$2 = r^2g_2 - rg_2 = (r-1)rg_2.\tag2$$
Dividing both sides of $(1)$ by both sides of $(2)$, we get
$$
\frac52 = \frac{r+1}{r}.
$$
So $r=2/3$.  And then from $(2)$, we get $g_2 = -9$.
And then $g_3=-6$ and $g_4=-4$.
