Use the nth term of a geometric sequence An=a1(r n-1) to answer the following questions 
1.What is the 5th term of the geometric sequence 3/20,3/2,15....?
2.Find the six term of a geometric sequence where the second term is 6 and the common ratio is 2.
3.Find K so that terms k-3,k+1 and 4k-2 form a geometric sequence.
4.In the geometric sequence 6,12,24,48,...., Which term is 768?
5.The second term of a geometric is 5/4 and it's fourth term is 

Let's solve each question step by step using the nth term of a geometric sequence formula: 

\[
A_n = a_1 \cdot r^{(n-1)}
\]

where \( A_n \) is the nth term, \( a_1 \) is the first term, and \( r \) is the common ratio.

### 1. What is the 5th term of the geometric sequence 3/20, 3/2, 15...?

**Step 1: Find the common ratio \( r \).**

The common ratio \( r \) is found by dividing the second term by the first term:

\[
r = \frac{\frac{3}{2}}{\frac{3}{20}} = \frac{3}{2} \times \frac{20}{3} = 10
\]

**Step 2: Use the nth term formula to find the 5th term.**

Here, \( a_1 = \frac{3}{20} \), \( r = 10 \), and \( n = 5 \).

\[
A_5 = \frac{3}{20} \times 10^{(5-1)} = \frac{3}{20} \times 10^4 = \frac{3}{20} \times 10000 = 1500
\]

So, the 5th term is **1500**.

### 2. Find the 6th term of a geometric sequence where the second term is 6 and the common ratio is 2.

**Step 1: Find the first term \( a_1 \).**

Given the second term (\( A_2 \)) is 6:

\[
A_2 = a_1 \cdot r^{(2-1)} = a_1 \cdot 2 = 6 \implies a_1 = 3
\]

**Step 2: Use the nth term formula to find the 6th term.**

Now, \( a_1 = 3 \), \( r = 2 \), and \( n = 6 \).

\[
A_6 = 3 \times 2^{(6-1)} = 3 \times 2^5 = 3 \times 32 = 96
\]

So, the 6th term is **96**.

### 3. Find \( k \) so that the terms \( k-3 \), \( k+1 \), and \( 4k-2 \) form a geometric sequence.

For these terms to form a geometric sequence, the ratio of consecutive terms must be equal:

\[
\frac{k+1}{k-3} = \frac{4k-2}{k+1}
\]

Cross-multiply to solve:

\[
(k+1)^2 = (k-3)(4k-2)
\]

Expand both sides:

\[
k^2 + 2k + 1 = 4k^2 - 14k + 6
\]

Simplify:

\[
0 = 3k^2 - 16k + 5
\]

Solve this quadratic equation using the quadratic formula:

\[
k = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \times 3 \times 5}}{2 \times 3} = \frac{16 \pm \sqrt{256 - 60}}{6} = \frac{16 \pm \sqrt{196}}{6} = \frac{16 \pm 14}{6}
\]

\[
k = \frac{30}{6} = 5 \quad \text{or} \quad k = \frac{2}{6} = \frac{1}{3}
\]

So, \( k = 5 \) or \( k = \frac{1}{3} \).

### 4. In the geometric sequence 6, 12, 24, 48,...., which term is 768?

**Step 1: Find the common ratio \( r \).**

\[
r = \frac{12}{6} = 2
\]

**Step 2: Use the nth term formula to find \( n \).**

\[
A_n = 6 \cdot 2^{(n-1)} = 768
\]

Solve for \( n \):

\[
2^{(n-1)} = \frac{768}{6} = 128 \implies 2^{(n-1)} = 2^7 \implies n - 1 = 7 \implies n = 8
\]

So, the 8th term is **768**.

### 5. The second term of a geometric sequence is \( \frac{5}{4} \) and the fourth term is... 

Let’s denote the first term as \( a_1 = a \), the common ratio as \( r \).

Given:

\[
A_2 = a \cdot r = \frac{5}{4}
\]

\[
A_4 = a \cdot r^3
\]

To proceed, please provide the fourth term to find the rest of the values.

Would you like further details or have any questions?

Here are some related questions you might consider:

1. What is the 7th term of a geometric sequence given the first term and common ratio?
2. How can we find the sum of the first n terms of a geometric sequence?
3. How does the common ratio affect the growth of a geometric sequence?
4. What happens if the common ratio is a fraction or negative?
5. How to determine if a sequence is geometric given a set of terms?

**Tip:** In a geometric sequence, each term is obtained by multiplying the previous term by a constant (the common ratio). Understanding the common ratio helps to quickly identify the pattern of the sequence!

Learn how to solve geometric sequence problems using the nth term formula. Includes examples such as finding specific terms and solving for unknowns in geometric sequences. Suitable for Grades 9-12 students.

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