For items 6 – 10, find the specified term of the following geometric sequences. 
6.) 3, 6, 12, … find a7
7.) 10, 5, 5
2
, … find a6
8.) 1, 3, 9, … find a9
9.) 2, – 1 , 1
2
, … find a10
10.) 100, 20, 4, … find a8
For items 11 – 15, insert geometric means in each geometric sequence. 
11.) 3, ____, 8
12.) 3
2
, _____, 
3
8
13.) 7, _____, _____, 56
14.) 36, _____, _____, _____, 9
64
15.) – 1, _____, – 9 , _____, – 81



ASSESSMENT 1. FIND OUT!
Find the specified term of each of the given geometric sequences. Show your complete 
solutions. 
1.) 3, 6, 12, … 𝑎7
2.) 18, 6, 2, … 𝑎9
3.) 1,
1
2
,
1
4
, … 𝑎7




ASSESSMENT 1: HALFWAY
Find the geometric mean of the given extremes. 
Given Geometric Mean
1. 3 and 8
2. 100 and 25
3. 1
2
 and 1
8
4. 3 and 1
3
5. x and x7
ASSESSMENT 2. IN BETWEEN
Insert geometric means in each geometric sequence. 
1.) 2, _____, _____, 686
2.) _____, 24, _____, _____, 3
64
3.) _____, _____, _____, 4, 8 
4.) _____, 
1
4
, 
1
2
, _____
5.) 81, _____, _____, _____, _____, 1



Let's break this down step by step:

### Items 6 – 10: Finding the Specified Term of a Geometric Sequence

For each geometric sequence, the nth term can be found using the formula:

\[
a_n = a_1 \times r^{(n-1)}
\]

Where:
- \( a_n \) is the nth term.
- \( a_1 \) is the first term.
- \( r \) is the common ratio.
- \( n \) is the term number.

Let's solve each problem.

#### 6. Sequence: 3, 6, 12, … Find \( a_7 \)
- \( a_1 = 3 \)
- \( r = \frac{6}{3} = 2 \)
- \( n = 7 \)

\[
a_7 = 3 \times 2^{(7-1)} = 3 \times 2^6 = 3 \times 64 = 192
\]

**Answer: \( a_7 = 192 \)**

#### 7. Sequence: 10, 5, \( \frac{5}{2} \), … Find \( a_6 \)
- \( a_1 = 10 \)
- \( r = \frac{5}{10} = \frac{1}{2} \)
- \( n = 6 \)

\[
a_6 = 10 \times \left(\frac{1}{2}\right)^{(6-1)} = 10 \times \left(\frac{1}{2}\right)^5 = 10 \times \frac{1}{32} = \frac{10}{32} = \frac{5}{16}
\]

**Answer: \( a_6 = \frac{5}{16} \)**

#### 8. Sequence: 1, 3, 9, … Find \( a_9 \)
- \( a_1 = 1 \)
- \( r = \frac{3}{1} = 3 \)
- \( n = 9 \)

\[
a_9 = 1 \times 3^{(9-1)} = 3^8 = 6561
\]

**Answer: \( a_9 = 6561 \)**

#### 9. Sequence: 2, –1, \( \frac{1}{2} \), … Find \( a_{10} \)
- \( a_1 = 2 \)
- \( r = \frac{-1}{2} = -\frac{1}{2} \)
- \( n = 10 \)

\[
a_{10} = 2 \times \left(-\frac{1}{2}\right)^{(10-1)} = 2 \times \left(-\frac{1}{2}\right)^9 = 2 \times -\frac{1}{512} = -\frac{2}{512} = -\frac{1}{256}
\]

**Answer: \( a_{10} = -\frac{1}{256} \)**

#### 10. Sequence: 100, 20, 4, … Find \( a_8 \)
- \( a_1 = 100 \)
- \( r = \frac{20}{100} = \frac{1}{5} \)
- \( n = 8 \)

\[
a_8 = 100 \times \left(\frac{1}{5}\right)^{(8-1)} = 100 \times \left(\frac{1}{5}\right)^7 = 100 \times \frac{1}{78125} = \frac{100}{78125} = \frac{4}{3125}
\]

**Answer: \( a_8 = \frac{4}{3125} \)**

### Items 11 – 15: Inserting Geometric Means

To insert geometric means, find the common ratio \( r \) by dividing the last term by the first and taking the appropriate root depending on the number of means to be inserted. Then multiply each preceding term by \( r \) to find the next term.

#### 11. Sequence: 3, ____, 8
- We need to find \( r \) such that \( 8 = 3r^2 \).

\[
r^2 = \frac{8}{3}, \quad r = \sqrt{\frac{8}{3}} = \frac{2\sqrt{6}}{3}
\]

The geometric mean is \( 3 \times \frac{2\sqrt{6}}{3} = 2\sqrt{6} \).

**Answer: \( 3, \, 2\sqrt{6}, \, 8

Learn how to find specified terms in geometric sequences and insert geometric means. This detailed guide covers various examples of geometric sequences and provides step-by-step solutions. Suitable for high school students.

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