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!
Best Answer
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 $g_3 = g_1r^2$.