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What is the tenth term of the geometric sequence that has a common ratio of [tex]$\frac{1}{3}$[/tex] and 36 as its fifth term?

A. [tex]$\frac{4}{27}$[/tex]
B. [tex][tex]$\frac{27}{4}$[/tex][/tex]
C. [tex]$\frac{4}{81}$[/tex]
D. [tex]$\frac{1}{36}$[/tex]


Sagot :

To find the tenth term of the geometric sequence that has a common ratio of [tex]\(\frac{1}{3}\)[/tex] and 36 as its fifth term, we need to follow these steps:

1. Identify the general formula of the geometric sequence:
The [tex]\(n\)[/tex]-th term [tex]\(a_n\)[/tex] of a geometric sequence can be expressed as:
[tex]\[ a_n = a_1 \cdot r^{n-1} \][/tex]
where:
- [tex]\(a_1\)[/tex] is the first term,
- [tex]\(r\)[/tex] is the common ratio,
- [tex]\(n\)[/tex] is the term number.

2. Given values:
[tex]\[ r = \frac{1}{3}, \quad a_5 = 36, \quad n = 10 \][/tex]

3. Find the first term [tex]\(a_1\)[/tex]:
Since [tex]\(a_5 = 36\)[/tex], we can use the general formula:
[tex]\[ a_5 = a_1 \cdot r^{5-1} = a_1 \cdot r^4 \][/tex]
Substituting the given values:
[tex]\[ 36 = a_1 \cdot \left(\frac{1}{3}\right)^4 \][/tex]
Solving for [tex]\(a_1\)[/tex]:
[tex]\[ 36 = a_1 \cdot \frac{1}{81} \implies a_1 = 36 \cdot 81 = 2916 \][/tex]

4. Calculate the tenth term [tex]\(a_{10}\)[/tex]:
Using the general formula again:
[tex]\[ a_{10} = a_1 \cdot r^{10-1} = a_1 \cdot r^9 \][/tex]
Substituting the values:
[tex]\[ a_{10} = 2916 \cdot \left(\frac{1}{3}\right)^9 \][/tex]
Simplifying:
[tex]\[ a_{10} = 2916 \cdot \frac{1}{19683} = \frac{2916}{19683} \][/tex]

5. Simplify the fraction:
[tex]\[ \frac{2916}{19683} = \frac{4}{81} \][/tex]

Hence, the tenth term of the geometric sequence is [tex]\(\frac{4}{81}\)[/tex].

The correct answer is:
C. [tex]\(\frac{4}{81}\)[/tex]