Math Problem Statement

A person decided to invest money at the end of each quarter. His initial investment amount is R1175 but thereafter, starting with the second quarter, he decreased the amount by R42 each time. The accumulated investment after 2 years if interest is compounded at i(4) = 18% p.a., is? Round off to two decimals only right at the end.

Solution

We are given a scenario where a person invests a decreasing amount of money at the end of each quarter, with the investment amount decreasing by R42 each quarter after the initial investment of R1175. The interest is compounded quarterly at a rate of i(4)=18%i(4) = 18\% p.a.

To find the accumulated investment after 2 years, let's break it down step by step.

Step 1: Define the terms

  • The initial investment at the end of the first quarter is R1=R1175R_1 = R1175.
  • The investment decreases by R42R42 per quarter, so the investment at the end of the second quarter will be R2=117542=1133R_2 = 1175 - 42 = 1133, and so on.
  • The interest is compounded quarterly at 18% per year, so the quarterly interest rate i=18%4=4.5%=0.045i = \frac{18\%}{4} = 4.5\% = 0.045.
  • The total number of quarters in 2 years is 2×4=82 \times 4 = 8.

This is an example of a decreasing annuity because the investment amount reduces by a fixed amount every quarter.

Step 2: Formula for the future value of a decreasing annuity

The future value FVFV of an annuity with decreasing payments can be calculated using the following formula:

FV=P1×(1+i)n1+P2×(1+i)n2++PnFV = P_1 \times (1 + i)^{n-1} + P_2 \times (1 + i)^{n-2} + \dots + P_n

Where:

  • PkP_k is the payment at the end of the kk-th quarter.
  • i=0.045i = 0.045 is the quarterly interest rate.
  • n=8n = 8 is the total number of payments (quarters).

Since the payments decrease by a fixed amount, Pk=1175(k1)×42P_k = 1175 - (k-1) \times 42.

Step 3: Calculate each term and the total future value

Let's calculate the payments and their future values one by one:

  1. P1=1175P_1 = 1175, compounded for 7 periods: 1175×(1+0.045)71175 \times (1 + 0.045)^7
  2. P2=1133P_2 = 1133, compounded for 6 periods: 1133×(1+0.045)61133 \times (1 + 0.045)^6
  3. P3=1091P_3 = 1091, compounded for 5 periods: 1091×(1+0.045)51091 \times (1 + 0.045)^5
  4. P4=1049P_4 = 1049, compounded for 4 periods: 1049×(1+0.045)41049 \times (1 + 0.045)^4
  5. P5=1007P_5 = 1007, compounded for 3 periods: 1007×(1+0.045)31007 \times (1 + 0.045)^3
  6. P6=965P_6 = 965, compounded for 2 periods: 965×(1+0.045)2965 \times (1 + 0.045)^2
  7. P7=923P_7 = 923, compounded for 1 period: 923×(1+0.045)1923 \times (1 + 0.045)^1
  8. P8=881P_8 = 881, no compounding (payment made at the end): 881881

Summing these values will give the total accumulated investment.

Step 4: Final Calculation

Now I will calculate each term and sum them to find the final future value.The accumulated investment after 2 years, with quarterly payments and interest compounded at 18% p.a., is approximately R9733.50.

Would you like further details or clarifications?

Here are five related questions to deepen your understanding:

  1. How does the formula change if the investment amount increases instead of decreasing?
  2. What happens to the accumulated amount if the interest rate is compounded monthly instead of quarterly?
  3. How would the future value change if the initial payment were higher?
  4. What if the interest rate was variable instead of fixed?
  5. How can we calculate the present value for the same investment plan?

Tip: When working with compound interest, always ensure you understand the compounding frequency as it significantly affects the final amount.

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Decreasing Annuities
Quarterly Compounding

Formulas

Future Value of a Decreasing Annuity: FV = P1 * (1 + i)^(n-1) + P2 * (1 + i)^(n-2) + ... + Pn
Quarterly Interest Rate: i = Annual Interest Rate / 4

Theorems

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Suitable Grade Level

Grades 10-12