The second term of an exponential magnitude is greater than the first term by 8. The sum of the second and third terms is 48. Show that there are two quality scales that meet these requirements and find

To solve the problem, we are given the following conditions for an exponential sequence:

1. The second term of the sequence is greater than the first term by 8.
2. The sum of the second and third terms is 48.

Let's define the terms of the exponential sequence as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)

Here, \( a \) is the first term, and \( r \) is the common ratio.

### Step 1: Write the equations based on the conditions.
From the first condition:
- The second term \( ar \) is greater than the first term \( a \) by 8. Therefore,
  \[
  ar = a + 8
  \]
  
From the second condition:
- The sum of the second term \( ar \) and the third term \( ar^2 \) is 48. Therefore,
  \[
  ar + ar^2 = 48
  \]

### Step 2: Solve for \( a \) in terms of \( r \).
Using the equation \( ar = a + 8 \), solve for \( a \):
\[
ar - a = 8
\]
\[
a(r - 1) = 8
\]
\[
a = \frac{8}{r - 1}
\]

### Step 3: Substitute \( a \) into the second equation.
Substitute \( a = \frac{8}{r - 1} \) into the equation \( ar + ar^2 = 48 \):
\[
\frac{8}{r - 1}r + \frac{8}{r - 1}r^2 = 48
\]
Factor out \( \frac{8r}{r - 1} \):
\[
\frac{8r(1 + r)}{r - 1} = 48
\]
Multiply both sides by \( r - 1 \):
\[
8r(1 + r) = 48(r - 1)
\]
Expand both sides:
\[
8r + 8r^2 = 48r - 48
\]
Move all terms to one side to form a quadratic equation:
\[
8r^2 - 40r + 48 = 0
\]
Simplify by dividing the entire equation by 8:
\[
r^2 - 5r + 6 = 0
\]

### Step 4: Solve the quadratic equation.
Solve the quadratic equation \( r^2 - 5r + 6 = 0 \) using factoring:
\[
r^2 - 5r + 6 = (r - 2)(r - 3) = 0
\]
So, \( r = 2 \) or \( r = 3 \).

### Step 5: Find the corresponding values of \( a \).
For \( r = 2 \):
\[
a = \frac{8}{2 - 1} = 8
\]

For \( r = 3 \):
\[
a = \frac{8}{3 - 1} = 4
\]

### Step 6: Verify the solutions.
- For \( r = 2 \) and \( a = 8 \), the sequence is \( 8, 16, 32 \). The second term is greater than the first by 8, and the sum of the second and third terms is \( 16 + 32 = 48 \).
- For \( r = 3 \) and \( a = 4 \), the sequence is \( 4, 12, 36 \). The second term is greater than the first by 8, and the sum of the second and third terms is \( 12 + 36 = 48 \).

Both conditions are satisfied for \( r = 2 \) and \( r = 3 \).

### Conclusion:
There are two possible sequences that meet the given conditions:
1. First term \( a = 8 \) and common ratio \( r = 2 \)
2. First term \( a = 4 \) and common ratio \( r = 3 \)

Would you like more details or have any questions?

Here are five related questions to explore:
1. What happens if the sum of the first and second terms is given instead?
2. How do you derive the general formula for the nth term in an exponential sequence?
3. Can the conditions of the problem be met if the terms are negative?
4. How would the solution change if the sum of the second and third terms were different?
5. What if the common ratio \( r \) were less than 1?

**Tip:** When solving problems involving exponential sequences, always start by expressing the terms in terms of the first term and the common ratio to simplify the relationships between them.

Explore how to solve exponential sequence problems involving the difference between the second and first terms, and the sum of the second and third terms. Detailed step-by-step solutions provided for grades 9-12.

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