Question

I need help answering this practice problem from my calculus prep book

Answer 1

We can use the following rule to write any term of a geometric sequence:

$a_n=a_1r^{(n-1)}$

Where a1 is the first term of the sequence and r is the common ratio.

We know that:

$\begin{gathered} a_1=32 \\ a_5=\frac{81}{8} \end{gathered}$

Now we can write the equation:

$\frac{81}{8}=32r^{(5-1)}$

Now we can solve:

$\begin{gathered} \frac{81}{8\cdot32}=r^4 \\ r=\sqrt[4]{\frac{81}{256}}=\frac{3}{4} \end{gathered}$

Now that we know the ratio, we can find any term in the sequence, by using:

$a_n=32\frac{3}{4}^{(n-1)}$

The problem ask us to find a2, a3 and a4:

$\begin{gathered} a_2=32\frac{3}{4}^{(2-1)}=24 \\ a_3=32\frac{3}{4}^{(3-1)}=18 \\ a_4=32\frac{3}{4}^{(4-1)}_{}=\frac{27}{2} \end{gathered}$