Equation of y, the population size after x hours is y = 1000 (1.15)ˣ
Further explanation
Firstly , let us learn about types of sequence in mathematics.
Arithmetic Progression is a sequence of numbers in which each of adjacent numbers have a constant difference.
[tex]\boxed{T_n = a + (n-1)d}[/tex]
[tex]\boxed{S_n = \frac{1}{2}n ( 2a + (n-1)d )}[/tex]
Tn = n-th term of the sequence
Sn = sum of the first n numbers of the sequence
a = the initial term of the sequence
d = common difference between adjacent numbers
Geometric Progression is a sequence of numbers in which each of adjacent numbers have a constant ration.
[tex]\boxed{T_n = a ~ r^{n-1}}[/tex]
[tex]\boxed{S_n = \frac{a( 1 - r^n ) }{1 - r}}[/tex]
Tn = n-th term of the sequence
Sn = sum of the first n numbers of the sequence
a = the initial term of the sequence
r = common ratio between adjacent numbers
Let us now tackle the problem!
Let :
Initial number of population = a
After 5 hours → Population = y = a × 2 = 2000
2a = 2000
a = 2000 ÷ 2
a = 1000
After 10 hours → Population = y = a × 2 × 2 = a × 2² = 1000 × 2²
After 15 hours → Population = y = a × 2 × 2 × 2 = a × 2³ = 1000 × 2³
After x hours ↓
Population = y = [tex]1000 \times 2^{\frac{x}{5}}[/tex]
Population = y = [tex]1000 \times (2^{\frac{1}{5}})^x[/tex]
Population = y = [tex]\large{ \boxed {1000 \times (1.15)^x} }[/tex]
Let's prove the equation above :
After 5 hours :
Population = y = [tex]1000 \times (1.15)^5[/tex] = 2011 ≈ 2000 ✔
After 10 hours :
Population = y = [tex]1000 \times (1.15)^{10}[/tex] = 4046 ≈ 4000 ✔
After 15 hours :
Population = y = [tex]1000 \times (1.15)^{15}[/tex] = 8137 ≈ 8000 ✔
Learn more
- Geometric Series : https://brainly.com/question/4520950
- Arithmetic Progression : https://brainly.com/question/2966265
- Geometric Sequence : https://brainly.com/question/2166405
Answer details
Grade: Middle School
Subject: Mathematics
Chapter: Arithmetic and Geometric Series
Keywords: Arithmetic , Geometric , Series , Sequence , Difference , Term
