Hurryyy a bacteria population doubles in size every 5 hours. after 5 hours, the sample contains 2,000 bacteria. which equation best models y, the population size after x hours? y = 1000(1.15)x y = 1000(2)x y = 1000(2.5)x y = 1000(3)x

Equation of y, the population size after x hours is y = 1000 (1.15)ˣ

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.

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.

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 =

Population = y =

Population = y =

Let's prove the equation above :

After 5 hours :

Population = y = = 2011 ≈ 2000 ✔

After 10 hours :

Population = y = = 4046 ≈ 4000 ✔

After 15 hours :

Population = y = = 8137 ≈ 8000 ✔

Grade: Middle School

Subject: Mathematics

Chapter: Arithmetic and Geometric Series

Keywords: Arithmetic , Geometric , Series , Sequence , Difference , Term

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}