A person just fitted for contact lenses is told to wear them only 2 hours the first day and to increase the amount of time by 20 minutes each day. After how many days will the person be able to wear the contacts for 14 hours?

*Solve using a arithmetic series

Well, let's calculate the number of days it will take. We can create an arithmetic series to represent the increasing amount of time the person wears the contacts each day.

The formula for an arithmetic series is: Sn = (n/2)(a + l), where Sn is the sum, n is the number of terms, a is the first term, and l is the last term.

In this case, the first term (a) is 2 hours, and we want to find the number of days (n) it will take to reach 14 hours as the last term (l).

Let's set up the equation: 14 = (n/2)(2 + 14).

Simplifying, we get: 14 = (n/2)(16).

Dividing both sides by 16, we have: 14/16 = n/2.

Simplifying further, we get: 7/8 = n/2.

Now, let's solve for n: 7 * 2 = 8 * n.

Thus, 14 = 8n.

Dividing both sides by 8, we find: n = 14/8.

So, after approximately 1.75 days, the person will be able to wear the contacts for 14 hours. However, since we can't have a fraction of a day, we'll round up to the nearest whole number. Therefore, it will take about 2 days for the person to be able to wear the contacts for 14 hours.

To solve this problem using an arithmetic series, we need to find the number of days it will take for the person to wear the contacts for 14 hours.

First, let's calculate the total number of hours the person wears the contacts each day:

Day 1: 2 hours
Day 2: 2 hours + 20 minutes = 2 hours and 20 minutes = 2.33 hours
Day 3: 2.33 hours + 20 minutes = 2.33 hours + 0.33 hours = 2.66 hours
Day 4: 2.66 hours + 20 minutes = 2.66 hours + 0.33 hours = 2.99 hours

We can see that the person adds 20 minutes each day, which is equivalent to 1/3 of an hour. So, the number of hours the person wears the contacts each day can be represented by an arithmetic series:

2, 2.33, 2.66, 2.99, ...

Now, let's find the number of terms in this series when the sum is equal to 14 hours. We can use the formula for the sum of an arithmetic series:

Sn = (n/2)(2a + (n-1)d)

where Sn is the sum of the series, a is the first term, d is the common difference, and n is the number of terms.

In this case, Sn = 14 hours, a = 2 hours, and d = 1/3 hours.

14 = (n/2)(2(2) + (n-1)(1/3))
14 = (n/2)(4 + (n-1)/3)

Multiplying by 2 to eliminate fractions:

28 = n(4 + (n-1)/3)
28 = 3n(4 + (n-1))
28 = 3n(4 + n-1)
28 = 3n(3 + n)

Expanding:

28 = 9n + 3n^2
3n^2 + 9n - 28 = 0

Using the quadratic formula to solve for n:

n = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 3, b = 9, and c = -28:

n = (-9 ± √(9^2 - 4(3)(-28))) / (2(3))
n = (-9 ± √(81 + 336)) / 6
n = (-9 ± √417) / 6

Taking the positive value:

n = (-9 + √417) / 6 ≈ 2.244

Since the number of days must be a whole number, we round up to the nearest whole number:

n ≈ 3

Therefore, it will take approximately 3 days for the person to be able to wear the contacts for 14 hours.