Question

Find the general function for a geometric sequence in which a5=2
and a8=1/4
.



an=1/8(2)^(n−1)


an=2(32)^(n−1)


an=32(1/2)^n


an=32(1/2)^(n−1)


an=(1/2)(32)^(n−1)

Answer 1

To solve for the general function of a geometric sequence given that \( a_5 = 2 \) and \( a_8 = \frac{1}{4} \), we start by using the property of geometric sequences.

In a geometric sequence, the \( n \)-th term can be represented as:
\[
a_n = a_1 r^{n-1}
\]
where \( a_1 \) is the first term and \( r \) is the common ratio.

We can express \( a_5 \) and \( a_8 \) in terms of \( a_1 \) and \( r \):
\[
a_5 = a_1 r^{5-1} = a_1 r^4 = 2
\]
\[
a_8 = a_1 r^{8-1} = a_1 r^7 = \frac{1}{4}
\]

Now we have the following two equations:
1. \( a_1 r^4 = 2 \) \( \quad (1) \)
2. \( a_1 r^7 = \frac{1}{4} \) \( \quad (2) \)

Next, we can divide equation (2) by equation (1):
\[
\frac{a_1 r^7}{a_1 r^4} = \frac{\frac{1}{4}}{2}
\]
This simplifies to:
\[
r^3 = \frac{1/4}{2} = \frac{1}{8}
\]

Taking the cube root of both sides gives us:
\[
r = \frac{1}{2}
\]

Now we substitute \( r \) back into equation (1) to find \( a_1 \):
\[
a_1 \left( \frac{1}{2} \right)^4 = 2
\]
This simplifies to:
\[
a_1 \cdot \frac{1}{16} = 2
\]
Multiplying both sides by 16, we get:
\[
a_1 = 32
\]

Now we can write the general formula for the geometric sequence:
\[
a_n = a_1 r^{n-1} = 32 \left( \frac{1}{2} \right)^{n-1}
\]
This can also be written as:
\[
a_n = 32 \cdot 2^{-(n-1)} = 32 \cdot \frac{1}{2^{n-1}}
\]

To match this with the provided options, we can rewrite it as:
\[
a_n = 32 \left( \frac{1}{2} \right)^{n-1}
\]

Thus, the correct option is:
\[
\boxed{a_n = 32 \left( \frac{1}{2} \right)^{n-1}}
\]