Math Lesson 12.2.7 - The Practical Applications of Number Series and Sequences

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Welcome to our Math lesson on The Practical Applications of Number Series and Sequences, this is the seventh lesson of our suite of math lessons covering the topic of Working with Arithmetic and Geometric Series. How to find the Sum of the First n-Terms of a Series., you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Understanding the Practical Applications of Number Series and Sequences

Number series and sequences are very common in practice. We can find them everywhere, from banking to construction; from physics to engineering and so on. Let's consider a couple of examples: one from each type of series discussed earlier. Let's begin with an example involving arithmetic series.

Example 10

Lisa bought a $12,000 car (including interests) from a car dealer shop. She pays for the car by making monthly payments that form an arithmetic sequence. The first payment is $250 and the debt is entirely repaid in 16 installments, with a total interest of 10%. Calculate:

1. The amount of the last installment to be paid.
2. The increase in payment Lisa has to pay in respect to the previous installment
3. The amount of the 10th installment

Solution 10

1. First, let's calculate the whole amount to be paid, which corresponds to the sum S_n of the arithmetic series obtained with the given numbers. Thus, since the total interest is 10%, a value that corresponds to 10/100 of $12,000 = $1200, the total amount Lisa has to pay (in USD) is
   S_n = 12,000 + 1,200
   = 13,200
   From the Gauss formula for the arithmetic series, we have
   S_n = (x₁ + x_n ) ∙ n/2
   Thus, given that x₁ = 250, n = 16 and S_n = 13,200, we obtain for the last installment x_n:
   13,200 = (250 + x_n ) ∙ 16/2
   250 + x_n = 13,200 ∙ 2/16
   250 + x_n = 1650
   x_n = 1650 - 250
   x_n = 1400
   Therefore, Lisa has to pay $1400 for her last installment.
2. To find the increase in the payment means to find the common difference d of the corresponding arithmetic series, where x₁ = 250, x_n = 1400 and n = 16. Thus,
   x_n = x₁ + (n - 1) ∙ d
   d = x_n - x₁/n - 1
   = 1400 - 250/16
   = 71.875
   Therefore, the installments increase by $71.875 every time.
3. To find the amount of the 10th installment means finding x₁₀ in the corresponding sequence (n = 10). Thus, since
   x_n = x₁ + (n - 1) ∙ d
   we obtain after substituting the known values
   x₁₀ = 250 + (10-1) ∙ 71.875
   = 250 + 9 ∙ 71.875
   = 896.875
   Therefore, Lisa has to pay $696.875 for her 10th installment.

Now, let's see another practical example, but this time with geometric series.

Example 11

Alex is a high school student. To make him more engaged in lessons, his father has promised to add an extra 20% to the monthly pocket money he gives to Alex for every 90 + points Alex scores in exams. The standard amount of monthly pocket money Alex received from his father before this agreement was $20. Calculate

1. The monthly pocket money Alex receives from his father if he scores 90 + points in 13 exams in that month.
2. The amount of money Alex receives in the last month if he increases his 90 + score by 2 exams every month during the entire academic year (9 months), given that in the first month he has scored 90 + in three exams.

Solution 12

1. This is an example of a geometric sequence (and series), with a common ratio R = 1.2 giving that the amount increases by 20% every time Alex scores 90 + point in an exam (100% + 20% = 120% = 120/100 = 1.2). The general term of this sequence is
   x_n = x₁ ∙ R^{n - 1}
   where x₁ = 200 and R = 1.2. This formula is valid is for one month only, as the values reset each month. Thus, the (monthly) general term for this specific case is
   x_n = 20 ∙ 1.2^{n - 1}
   Thus, if Alex scores 90 + points in 13 exams in a certain month (n = 13), he will receive
   x₁3 = 20 ∙ 1.2^{13 - 1}
   = 20 ∙ 1.2¹²
   = 20 ∙ 8.9161
   = $178.322
   in that month.
2. The number of exams Alex scores 90 + points forms an arithmetic sequence, where the first term is y₁ = 3 and the common difference is d = 2. Thus, given that the agreement is valid for an entire academic year which lasts 9 months (n = 9), we obtain for the last term of this sequence that corresponds to the number of 90 + scores Alex obtains in the last month:
   y_n = y₁ + (n - 1) ∙ d
   = 3 + (9 - 1) ∙ 2
   = 3 + 8 ∙ 2
   = 3 + 16
   = 19
   This number corresponds to the value of n in the geometric sequence that gives the monthly pocket money Alex receives. Hence, in the last month he will receive
   x_n = x₁ ∙ R^{n - 1}
   = 20 ∙ 1.2^{19 - 1}
   = 20 ∙ 1.2¹⁸
   = 20 ∙ 26.62
   = $532.4

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