Geometric Sequence Phrase Issues

This lesson will present you remedy a wide range of geometric sequence phrase issues.

Instance #1:

The inventory’s value of an organization isn’t doing effectively currently. Suppose the inventory’s value is 92% of its earlier value every day. What’s the inventory’s value after 10 days if the inventory was value $2500 proper earlier than it began to go down?  

Resolution

To resolve this drawback, we want the geometric sequence system proven beneath.

a_n = a₁ × r^{(n – 1)}

a₁ = authentic worth of the inventory  = 2500

a₂ = worth of the inventory after 1 day

a₁₁ = worth of the inventory after 10 days

r = 0.92

a₁₁ = 2500 × (0.92)^{(11 – 1)}

a₁₁ = 2500 × (0.92)¹⁰

a₁₁ = 2500 × 0.434

a₁₁ = $1085

The inventory’s value is about 1085 {dollars}.

Instance #2:

The third time period of a geometrical sequence is 45 and the fifth time period of the geometric sequence is 405. If all of the phrases of the sequence are optimistic numbers, discover the fifteenth time period of the geometric sequence.

Resolution

To resolve this drawback, we want the geometric sequence system proven beneath.

a_n = a₁ × r^{(n – 1)}

Discover the third time period

a₃ = a₁ × r^{(3 – 1)}

a₃ = a₁ × r²

Because the third time period is 45,  45 = a₁ × r² (equation 1)

Discover the fifth time period

a₅ = a₁ × r^{(5 – 1)}

a₅ = a₁ × r⁴

Because the fifth time period is 405,  405 = a₁ × r⁴ (equation 2)

Divide equation 2 by equation 1.

(a₁ × r⁴) / (a₁ × r²) = 405 / 45

Cancel a₁ since it’s each on prime and on the backside of the fraction.

r⁴ / r² = 9

r² = 9

r = ±√9

r = ±3

Use r  = 3, and equation 1 to discover a₁

45 = a₁ × (3)²

45 = a₁ × 9

a₁ = 45 / 9 = 5

a_n = a₁ × r^{(n – 1)}

Allow us to now discover a₁₅

a₁₅ = 5 × (3)^{(15 – 1)}

a₁₅ = 5 × (3)¹⁴ 

a₁₅ = 5 × 4782969 

a₁₅ =  23914845

Difficult geometric sequence phrase issues

Instance #3:

Suppose that the magnification of a PDF file on a desktop laptop is elevated by 15% for every degree of zoom. Suppose additionally that the unique size of the phrase “January” is 1.2 cm. Discover the size of the phrase “January” after 6 magnifications.

Resolution

To resolve this drawback, we want the geometric sequence system proven beneath.

a_n = a₁ × r^{(n – 1)}

a₁ = authentic size of the phrase  = 1.2 cm

a₂ = size of the phrase after 1 magnification

a₇ = size of the phrase 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 size of the phrase “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

Instance #4

Suppose that you really want a decreased copy of {a photograph}. The precise size of the {photograph} is 10 inches. If every discount is 64% of the unique, what number of reductions, will shrink the {photograph} to 1.07 inches.

Resolution

a_n = a₁ × r^{(n – 1)}

a₁ = authentic size of the {photograph}  = 10 inches

a₂ = size of the {photograph} after 1 discount

a_n = 1.07

r = 0.64

n = variety of reductions = ?

1.07 = 10 × (0.64)^{(n – 1)}

Divide either side by 10

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

0.107 = (0.64)^{(n – 1)}

Take the pure log of either side of the equation.

ln(0.107) = ln[(0.64)^{(n – 1)}]

Use the energy property of logarithms.

ln(0.107) = (n – 1)ln(0.64)

Divide either side 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 seek out ln(0.107) and ln(0.64)

n – 1 = -2.23492644452 -0.44628710262

n – 1 = 5.0078

n = 1 + 5.0078

n = 6.0078

Subsequently, you will want 6 reductions.