Find the third element of a geometric sequence who fifth element is 81 and whose ninth element is 16. Help please!
Hi reiward,
You should have been given the formula for finding the nth term of a geometric sequence.
$\displaystyle a_n=a_1 \cdot r^{n-1}$
$\displaystyle r$ = common ratio
Your first element ($\displaystyle a_1$) will be 81.
$\displaystyle n = 5$ because there are five elements in all between the 5th element and the 9th element, inclusive.
$\displaystyle a_n = 16$
Solve for $\displaystyle r$
Now that you have found the common ratio, substitute back into the original formula and solve for $\displaystyle a_3$.
This time there are 3 elements from the 3rd to the 5th elements, inclusive. So, $\displaystyle n=3$
Remember $\displaystyle a_5=81$
$\displaystyle a_5=a_3 \cdot r^{3-1}$
Hello, reiward!
Use the formula for the $\displaystyle n^{th}$ term: .$\displaystyle a_n \:=\:ar^{n-1}$
We are given: .$\displaystyle \begin{array}{ccccccc}a_5 &=& ar^4 &=& 81 & [1] \\ a_9 &=& ar^8 &=& 16 & [2] \end{array}$Find the third element of a geometric sequence who fifth element is 81
and whose ninth element is 16.
Divide [2] by [1]: .$\displaystyle \frac{ar^8}{ar^4} \:=\:\frac{16}{81} \quad\Rightarrow\quad r^4 \:=\:\frac{16}{81} \quad\Rightarrow\quad r \:=\:\pm\frac{2}{3}$
Substitute into [1]: .$\displaystyle a\left(\pm\frac{2}{3}\right)^4 \:=\:81 \quad\Rightarrow\quad \frac{16}{81}a\:=\:81 \quad\Rightarrow\quad a \:=\:\frac{6561}{16}$
Therefore: .$\displaystyle a_3\;=\;ar^2 \;=\;\frac{6561}{16}\left(\pm\frac{2}{3}\right)^2 \;=\;\frac{729}{4}$