Question

the table below represents a portion of a geometric sequence.
n | 1 | 2 | 3 | 4 | 5 | 6
f(n) | 4 | | | | | 128
enter the missing terms in order separated by a comma.

Explanation:

Step1: Recall geometric sequence formula

The general formula for a geometric sequence is $f(n) = a_1 r^{n-1}$, where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the term number. We know $a_1 = 4$ and $f(6)=128$.

Step2: Solve for common ratio $r$

Substitute $n=6$, $a_1=4$, $f(6)=128$ into the formula:
$$128 = 4 \times r^{6-1}$$
Simplify: $r^5 = \frac{128}{4} = 32$
Solve for $r$: $r = \sqrt[5]{32} = 2$

Step3: Calculate missing terms

• For $n=2$: $f(2)=4 \times 2^{2-1} = 4 \times 2 = 8$
• For $n=3$: $f(3)=4 \times 2^{3-1} = 4 \times 4 = 16$
• For $n=4$: $f(4)=4 \times 2^{4-1} = 4 \times 8 = 32$
• For $n=5$: $f(5)=4 \times 2^{5-1} = 4 \times 16 = 64$

Answer:

8, 16, 32, 64