Geometric sequence word problems

This lesson will show you how to solve a variety of geometric sequence word problems.

Example #1:

The stock's price of a company is not doing well lately. Suppose the stock's price is 92% of its previous price each day. What is the stock's price after 10 days if the stock was worth $2500 right before it started to go down?  

Solution

To solve this problem, we need the geometric sequence formula shown below.

a_n = a₁ × r^{(n - 1)}

a₁ = original value of the stock  = 2500

a₂ = value of the stock after 1 day

a₁₁ = value of the stock after 10 days

r = 0.92

a₁₁ = 2500 × (0.92)^{(11 - 1)}

a₁₁ = 2500 × (0.92)¹⁰

a₁₁ = 2500 × 0.434

a₁₁ = $1085

The stock's price is about 1085 dollars.

Example #2:

The third term of a geometric sequence is 45 and the fifth term of the geometric sequence is 405. If all the terms of the sequence are positive numbers, find the 15th term of the geometric sequence.

Solution

To solve this problem, we need the geometric sequence formula shown below.

a_n = a₁ × r^{(n - 1)}

Find the third term

a₃ = a₁ × r^{(3 - 1)}

a₃ = a₁ × r²

Since the third term is 45,  45 = a₁ × r² (equation 1)

Find the fifth term

a₅ = a₁ × r^{(5 - 1)}

a₅ = a₁ × r⁴

Since the fifth term is 405,  405 = a₁ × r⁴ (equation 2)

Divide equation 2 by equation 1.

(a₁ × r⁴) / (a₁ × r²) = 405 / 45

Cancel a₁ since it is both on top and at the bottom of the fraction.

r⁴ / r² = 9

r² = 9

r = ±√9

r = ±3

Use r  = 3, and equation 1 to find a₁

45 = a₁ × (3)²

45 = a₁ × 9

a₁ = 45 / 9 = 5

a_n = a₁ × r^{(n - 1)}

Let us now find a₁₅

a₁₅ = 5 × (3)^{(15 - 1)}

a₁₅ = 5 × (3)¹⁴ 

a₁₅ = 5 × 4782969 

a₁₅ =  23914845

Challenging geometric sequence word problems

Example #3:

Suppose that the magnification of a PDF file on a desktop computer is increased by 15% for each level of zoom. Suppose also that the original length of the word "January" is 1.2 cm. Find the length of the word "January" after 6 magnifications.

Solution

To solve this problem, we need the geometric sequence formula shown below.

a_n = a₁ × r^{(n - 1)}

a₁ = original length of the word  = 1.2 cm

a₂ = length of the word after 1 magnification

a₇ = length of the word after 6 magnifications

r = 1 + 0.15 = 1.15

n = 7

a₇ = 1.2 × (1.15)^{(7 - 1)}

a₇ = 1.2 × (1.15)⁶

a₇ = 1.2 × (1.15)⁶

a₇ = 1.2 × 2.313

a₇ = 2.7756

After 6 magnifications, the length of the word "January" is 2.7756 cm.

Day 1: a₁ = 1.2

Day 2: a₂ = 1.2 + 1.2(0.15) = 1.2(1 + 0.15)

Day 3: a₃ = 1.2(1 + 0.15) + [1.2(1 + 0.15)]0.15 = 1.2(1 + 0.15)(1 + 0.15) = 1.2(1 + 0.15)²

Day 7: a₇ = 1.2(1 + 0.15)⁶

Example #4

Suppose that you want a reduced copy of a photograph. The actual length of the photograph is 10 inches. If each reduction is 64% of the original, how many reductions, will shrink the photograph to 1.07 inches.

Solution

a_n = a₁ × r^{(n - 1)}

a₁ = original length of the photograph  = 10 inches

a₂ = length of the photograph after 1 reduction

a_n = 1.07

r = 0.64

n = number of reductions = ?

1.07 = 10 × (0.64)^{(n - 1)}

Divide both sides by 10

1.07 / 10 = [10 × (0.64)^{(n - 1)}] / 10

0.107 = (0.64)^{(n - 1)}

Take the natural log of both sides of the equation.

ln(0.107) = ln[(0.64)^{(n - 1)}]

Use the power property of logarithms.

ln(0.107) = (n - 1)ln(0.64)

Divide both sides of the equation by ln(0.64)

ln(0.107) / ln(0.64) = (n - 1)ln(0.64) / ln(0.64)

n - 1 = ln(0.107) / ln(0.64)

Use a calculator to find ln(0.107) and ln(0.64)

n - 1 = -2.23492644452 \ -0.44628710262

n - 1 = 5.0078

n = 1 + 5.0078

n = 6.0078

Therefore, you will need 6 reductions.