Patterns and Sequences Study Guide

Financial Mathematics Glossary

  • Appreciation: The increase in the value of an asset (such as a house, land, or investment) over time. This frequently occurs due to factors including demand, improvements, or inflation. Examples include land and real estate.
  • Depreciation: The decrease in the value of an asset over time. This typically happens due to usage, aging, or becoming outdated. An example is the value of a car.
  • Simple Interest: Interest that is computed solely on the principal and then added into it.
  • Compound Interest: Interest computed on both the principal and on the accumulated past interest.
  • Loan: Money borrowed from a bank or lender that must be paid back over time, usually with added interest.
  • Mortgage: A specific type of loan used to purchase real estate (land or a house). The property itself serves as a guarantee (collateral) until the loan is fully paid.
  • Investment: Putting money into something—like a business, stock, or savings plan—with the goal of earning more money in the future.
  • Accumulated Value (or Future Value): The total worth of an investment after a certain time, encompassing the original investment and all growth such as appreciation or interest.
  • Frequency of Conversion: The number of conversion periods occurring within one year.
  • Rate: The yearly rate of increase of the investment.
  • Period Rate: The rate of interest applicable to one single conversion period.
  • Time of Term: The duration, expressed in years, for which money is borrowed or invested.

Arithmetic Sequences and Series

Arithmetic Sequence Formula

\na_n = a_1 + d(n - 1)\n

Arithmetic Series (Sum) Formula

\nS_n = \frac{n}{2}(a_1 + a_n)\n

Example 1: Fixed Salary Increases

Mia was hired as an office assistant after graduating. In her 3rd year of employment, her monthly salary was ₱19,000₱19,000. The company provides a fixed salary increase of ₱1,500₱1,500 every year.

Finding the First Year Salary (a1a_1)

  • Given: a3=19,000a_3 = 19,000, d=1,500d = 1,500, n=3n = 3
  • Calculation:

\n19,000 = a_1 + 1,500(3 - 1)\n

\n19,000 - 3,000 = a_1\n

\na_1 = ₱16,000\n

  • Conclusion: Mia's first monthly salary is ₱16,000₱16,000.

Finding the Monthly Salary in the 8th Year (a8a_8)

  • Given: a1=16,000a_1 = 16,000, d=1,500d = 1,500, n=8n = 8
  • Calculation:

\na_8 = 16,000 + 1,500(8 - 1)\n

\na_8 = ₱26,500\n

  • Conclusion: Mia’s monthly salary in her 8th year of working is ₱26,500₱26,500.

Total Salary Received from the 1st to the 8th Year

  • Given:
    • First year annual salary (a1a_1): 16,000×12=192,00016,000 \times 12 = 192,000
    • Eight year annual salary (a8a_8): 26,500×12=318,00026,500 \times 12 = 318,000
    • n=8n = 8; d=1,500d = 1,500
  • Calculation:

\nS_8 = \frac{8}{2}(192,000 + 318,000)\n

\nS_8 = ₱2,040,000\n

  • Conclusion: Mia has earned a total of ₱2,040,000₱2,040,000 during her first 8 years in the company.
Example 2: Laptop Depreciation

Ana bought a laptop worth ₱45,000₱45,000. Its value decreases by ₱3,000₱3,000 every year.

  • Given: a1=45,000a_1 = 45,000, d=−3,000d = -3,000, n=5n = 5
  • Calculation for Value after 5 years (ana_n):

\na_5 = 45,000 - 3,000(5 - 1)\n

\na_5 = ₱33,000\n

  • Conclusion: The value of Ana’s laptop is ₱16,000₱16,000 after 5 years.

Geometric Sequences and Series

Geometric Sequence Formula

\na_n = a_1(r^{n-1})\n

Geometric Series (Sum) Formula

\nS_n = \frac{a_1(1-r^n)}{1-r}\n

Example 3: Annual Investment Growth

Ella invested money in a savings account earning 5%5\% interest annually. During the 4th year, the value of the investment was ₱57,881.25₱57,881.25.

Finding the Initial Investment (a1a_1)

  • Given: a4=57,881.25a_4 = 57,881.25, r=1.05r = 1.05, n=4n = 4
  • Calculation:

\n57,881.25 = a_1(1.05^{4-1})\n

\na_1 = \frac{57,881.25}{1.05^3}\n

\na_1 = ₱50,000\n

Value of Investment at the End of the 8th Year

  • Given: a1=50,000a_1 = 50,000, r=1.05r = 1.05, n=8n = 8
  • Calculation:

\na_8 = 50,000(1.05^{8-1})\n

\na_8 = ₱70,355.02\n

  • Conclusion: Ella has ₱70,355.02₱70,355.02 in her savings account at the end of the 8th year.

Sum of Annual Investment Values (1st through 8th Year)

  • Given: a1=50,000a_1 = 50,000, r=1.05r = 1.05, n=8n = 8
  • Calculation:

\nS_8 = 50,000\left(\frac{1 - 1.05^8}{1 - 1.05}\right)\n

\nS_8 = ₱477,455.44\n

  • Conclusion: The sum of Ella’s savings investment for 8 years is ₱477,455.44₱477,455.44.
Example 4: School Uniform Resale Value

A Grade 11 student bought a school uniform worth ₱2,500₱2,500. After each school year, the resale value decreases by 20%20\% due to wear and tear.

  • Given: a1=2,500a_1 = 2,500, r=1−0.2=0.8r = 1 - 0.2 = 0.8, n=4n = 4
  • Calculation:

\na_4 = 2,500(0.8^{4-1})\n

\na_4 = ₱1,280\n

  • Conclusion: The uniform value will only be ₱1,280₱1,280 after 4 years.

Compound Interest

Definition Compound interest is the interest earned on both the money originally invested (the principal) and the interest that has already been added over time.

The Compound Interest Formula

\nA = P(1 + i)^n\n

  • A = Accumulated value
  • P = Principal amount
  • i = Interest rate per period
  • n = Number of periods
Similarities with Geometric Sequences
  • Formula Structure:
    • Geometric Sequence: an=a1(rn−1)a_n = a_1(r^{n-1})
    • Compound Interest: A=P(1+i)nA = P(1 + i)^n
  • Accumulated Value: In geometric sequences, this is the term ana_n; in compound interest, this is AA.
  • Starting Amount: In geometric sequences, this is the first term a1a_1; in compound interest, this is the Principal amount PP.
  • Ratio/Growth Rate: The common ratio rr corresponds to the factor (1+i)(1 + i).
  • Timesteps: In geometric sequences, the first term typically starts at n=1n = 1 (making the exponent n−1n-1 for the nthn^{th} term). In compound interest, the principal amount starts at n=0n = 0 (making the exponent nn after nn periods).
Example 5: Student Loan Calculation

A Grade 12 student takes out a ₱50,000₱50,000 loan to buy a laptop for college. The loan charges 7.5%7.5\% compound interest per year. No payments are made during the first 3 years.

  • Given: P=50,000P = 50,000, i=0.075i = 0.075, n=3n = 3
  • Calculation:

\nA = 50,000(1 + 0.075)^3\n

\nA = ₱62,114.84\n

  • Conclusion: The student will have to pay ₱62,114.84₱62,114.84 at the end of 3 years.
Example 6: Mr. Santos Time Deposit

Mr. Santos invested Php15,000Php15,000 in a time deposit account that earns 6%6\% annual interest, compounded yearly. He plans to leave the money for 4 years.

  • Given: P=15,000P = 15,000, i=0.06i = 0.06, n=4n = 4
  • Calculation:

\nA = 15,000(1 + 0.06)^4\n

\nA = ₱18,937.15\n

  • Conclusion: Mr. Santos will have Php18,937.15Php18,937.15 after four years.