Math Problem Statement

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

Solution

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: aa
  • Second term: arar
  • Third term: ar2ar^2

Here, aa is the first term, and rr is the common ratio.

Step 1: Write the equations based on the conditions.

From the first condition:

  • The second term arar is greater than the first term aa by 8. Therefore, ar=a+8ar = a + 8

From the second condition:

  • The sum of the second term arar and the third term ar2ar^2 is 48. Therefore, ar+ar2=48ar + ar^2 = 48

Step 2: Solve for aa in terms of rr.

Using the equation ar=a+8ar = a + 8, solve for aa: ara=8ar - a = 8 a(r1)=8a(r - 1) = 8 a=8r1a = \frac{8}{r - 1}

Step 3: Substitute aa into the second equation.

Substitute a=8r1a = \frac{8}{r - 1} into the equation ar+ar2=48ar + ar^2 = 48: 8r1r+8r1r2=48\frac{8}{r - 1}r + \frac{8}{r - 1}r^2 = 48 Factor out 8rr1\frac{8r}{r - 1}: 8r(1+r)r1=48\frac{8r(1 + r)}{r - 1} = 48 Multiply both sides by r1r - 1: 8r(1+r)=48(r1)8r(1 + r) = 48(r - 1) Expand both sides: 8r+8r2=48r488r + 8r^2 = 48r - 48 Move all terms to one side to form a quadratic equation: 8r240r+48=08r^2 - 40r + 48 = 0 Simplify by dividing the entire equation by 8: r25r+6=0r^2 - 5r + 6 = 0

Step 4: Solve the quadratic equation.

Solve the quadratic equation r25r+6=0r^2 - 5r + 6 = 0 using factoring: r25r+6=(r2)(r3)=0r^2 - 5r + 6 = (r - 2)(r - 3) = 0 So, r=2r = 2 or r=3r = 3.

Step 5: Find the corresponding values of aa.

For r=2r = 2: a=821=8a = \frac{8}{2 - 1} = 8

For r=3r = 3: a=831=4a = \frac{8}{3 - 1} = 4

Step 6: Verify the solutions.

  • For r=2r = 2 and a=8a = 8, the sequence is 8,16,328, 16, 32. The second term is greater than the first by 8, and the sum of the second and third terms is 16+32=4816 + 32 = 48.
  • For r=3r = 3 and a=4a = 4, the sequence is 4,12,364, 12, 36. The second term is greater than the first by 8, and the sum of the second and third terms is 12+36=4812 + 36 = 48.

Both conditions are satisfied for r=2r = 2 and r=3r = 3.

Conclusion:

There are two possible sequences that meet the given conditions:

  1. First term a=8a = 8 and common ratio r=2r = 2
  2. First term a=4a = 4 and common ratio r=3r = 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 rr 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.

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Math Problem Analysis

Mathematical Concepts

Exponential Sequences

Formulas

Exponential sequence terms

Theorems

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Suitable Grade Level

Grades 9-12