APPLICATIONS OF GEOMETRIC SEQUENCES

Example 1 : 

A model predicts that the number of students taking the IB in a region will increase by 5% each year. If the number of IB students in the region is currently 12000, predict the number taking the IB in 5 years’ time.

Solution : 

Number of IB students in the region after 

1 year : 12000 ⋅ 1.05

2 years : 12000 ⋅ 1.05²

3 years : 12000 ⋅ 1.05³

This is a geometric sequence with 

a₁ = 12000 ⋅ 1.05 and r = 1.05

Use the formula a_n = a₁r^n-1 to predict the number of students taking the IB in 5 years’ time.

a₅ = 12000 ⋅ 1.05 ⋅ 1.05^{5 - 1}

= 12000 ⋅ 1.05 ⋅ 1.05⁴

= 12000 ⋅ 1.05⁵

≈ 15315

Example 2 :

The number of rabbits in a certain population doubles every month. If the population starts with 12 rabbits, what will the population of rabbits be 10 months from now?

Solution :

The population of rabbits after 

1 month = 12(2)

2 months = 12(2)²

3 months = 12(2)³

This is a geometric sequence with a₁ = 12(2) and r = 2.

Use the formula a_n = a₁r^n-1 to find the population of rabbits 10 months from now

a₁₀ = 12(2)(2)^{10 - 1}

= 12(2)(2)⁹

= 12(2)¹⁰

= 12(1024)

= 12,288

Example 3 :

The half of carbon-14 decays approximately in 6000 years. How much of 800 grams of this substance will remain after 30,000 years?

Solution :

One unit of time = 6000 years.

Amount of carbon-14 will remain after

1 unit of time : 800(0.5)

2 units of time : 800(0.5)²

3 units of time : 800(0.5)³

This is a geometric sequence with a₁ = 800(0.5) and r = 0.5.

30,000 years : 30000/6000 = 5 units of time.  

Use the formula a_n = a₁r^n-1 to find the amount of carbon-14 will remain after 30,000 years (5 units of time).

a₅ = 800(0.5)(0.5)^{5 - 1}

= 800(0.5)(0.5)⁴

= 800(0.5)⁵

= 25 grams

Example 4 : 

The present value of a machine is $40,000 and its value depreciates each year by 10%. Find the estimated value of the machine in the 6^th year.

Solution : 

Value of machine in the

1^st year : 40000

2^nd year : 40000 ⋅ 0.9

3^rd year : 40000 ⋅ 0.9²

This is a geometric sequence with 

a₁ = 4000 and r = 0.9

Use the formula a_n = a₁r^n-1 the estimated value of the machine in the 6^th year.

a₆ = 40000 ⋅ 0.9^{6 - 1}

= 40000 ⋅ 0.9⁵

= $23,619.60

Example 5 : 

Jake is generally selfish and somewhat unpopular at school. HE decided that he could improve his image by sharing his candy with people at school. When he has a pile of 1, 048, 576 candies, he generously plans to give away 75% of the candies that in the pile each day. Although Jake may be earning more candies for doing his homework, he is only giving away candies from the pile that started with 1, 048, 576.

a) How many candies will be in the pile after 8 days giving candy away ?

Solution : 

Number of candies initially = 1, 048, 576

Number of candies after 1 day = 0.25% of 1048576

= 262144

Number of candies after second day = 0.25% of 262144

= 65536

1048576, 262144, 65536, ..............

First term = 262144, common ratio = 0.25

a_n = ar^n-1

When n = 8

a)

a₈ = 262144(0.25)^8-1

= 262144(0.25)⁷

= 262144(0.0000610)

= 16 candies

Example 6 : 

A geometric series has a sum of 1365. Each term increases by a factor of 4. If there are 6 terms, find the value of the first term.

Solution :

S_n = 1365

Common difference = 4

Number of terms = 6

Then, S₆ = 1365

S_n = a(rⁿ - 1)/(r - 1)

a(4⁶ - 1)/(4 - 1) = 1365

a(4⁶ - 1)/(4 - 1) = 1365

a(4096 - 1)/3 = 1365

a(4095/3) = 1365

a(1365) = 1365

a = 1365/1365

a = 1

So, the first term is 1.

Example 7 : 

Clicking the zoom-out button on a mapping website doubles the side length of the square map. After how many clicks on the zoom-out button is the side length of the map 640 miles?

Solution :

a = 5, r = t₂/t₁

r = 10/5

r = 2

After how many clicks on the zoom out button is the side length of the map = 640 miles.

a_n = ar^n-1

640 = 5(2)^n-1

640/5 = 2^n-1

128 = 2^n-1

2⁷ = 2^n-1

Since bases are equal, we can equate the powers. 

n - 1 = 7

n = 8

So, the required number of clicks is 8.

Example 8 : 

A badminton tournament begins with 128 teams. After the first round, 64 teams remain. After the second round, 32 teams remain. How many teams remain after the third, fourth, and fifth rounds?

Solution :

Number of teams at the beginning = 128

Number of teams after the first round = 64

Number of teams after the second round = 32

128, 64, 32, ..............

a = 128

r = 64/128

r = 1/2

a_n = ar^n-1

Number of teams after the third round :

a₄ = 128(1/2)^4-1

= 128 (1/8)

= 16

Number of teams after the fourth round :

a₅ = 128(1/2)^5-1

= 128 (1/16)

= 8

Number of teams after the fifth round :

a₆ = 128(1/2)^6-1

= 128 (1/32)

= 4

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