# Compound Interest with Periodic Deductions

We will learn how to calculate the compound interest with periodic deductions or additions to the amount.

Solved examples on compound interest with periodic deductions:

1. Ron borrows $10,000 at a compound interest rate of 8% per annum. If he repays$ 2000 at the end of each year, find the sum outstanding at the end of the third year.

Solution:

For the first year:

Principal = $10,000 Rate = 8 % Time = 1 year Therefore, interest =$$$\frac{P × R × T}{100}$$

= $$$\frac{10000 × 8 × 1}{100}$$ =$$$\frac{80000}{100}$$

= $800 Therefore, the amount of loan after 1 year = Principal + Interest =$ 10,000 + $800 =$ 10,800

Ron pays back $2,000 at the end of the first year. So, the new principal at the beginning of the second year =$ 10,800 - $2,000 =$ 8,800

Therefore, for the second year:

Principal = $8,800 Rate = 8 % Time = 1 year Therefore, interest =$$$\frac{P × R × T}{100}$$

= $$$\frac{8,800 × 8 × 1}{100}$$ =$$$\frac{70400}{100}$$

= $704 Therefore, the amount of loan after 2 year = Principal + Interest =$ 8,800 + $704 =$ 9504

Ron pays back $2,000 at the end of the second year. So, the new principal at the beginning of the third year =$ 9504 - $2,000 =$ 7504

Therefore, for the third year:

Principal = $7504 Rate = 8 % Time = 1 year Therefore, interest =$$$\frac{P × R × T}{100}$$

= $$$\frac{7504 × 8 × 1}{100}$$ =$$$\frac{60032}{100}$$

= $600.32 Therefore, the amount of loan (outstanding sum) after 3 year = Principal + Interest =$ 7504 + $600.32 =$ 8104.32

2. Davis invests $20,000 at the beginning of every year in a bank and earns 10 % annual interest, compounded at the end of the year. What will be his balance in the bank at the end of three years. Solution: For the first year: Principal =$ 20,000

Rate = 10 %

Time = 1 year

Therefore, interest = $$$\frac{P × R × T}{100}$$ =$$$\frac{20000 × 10 × 1}{100}$$

= $$$\frac{200000}{100}$$ =$ 2000

Therefore, the amount at the end of the 1 year = Principal + Interest

= $20,000 +$ 2000

= $22,000 Davis deposits$ 20,000 at the beginning of the second year.

So, the new principal for the second year = $22,000 +$ 20,000

= $42,000 Therefore, for the second year: Principal =$ 42,000

Rate = 10 %

Time = 1 year

Therefore, interest = $$$\frac{P × R × T}{100}$$ =$$$\frac{42000 × 10 × 1}{100}$$

= $$$\frac{420000}{100}$$ =$ 4,200

Therefore, the amount at the end of the 2 year = Principal + Interest

= $42,000 +$ 4,200

= $46,200 Davis deposits$ 20,000 at the beginning of the third year.

So, the new principal for the third year = $46,200 +$ 20,000

= $66,200 Therefore, for the third year: Principal =$ 66,200

Rate = 10 %

Time = 1 year

Therefore, interest = $$$\frac{P × R × T}{100}$$ =$$$\frac{66200 × 10 × 1}{100}$$

= $$$\frac{662000}{100}$$ =$ 6620

Therefore, the amount at the end of the 3 year = Principal + Interest

= $66,200 +$ 6,620

= $72,820 Therefore, the balance in the bank at the end of thee years will be$ 72,820.

From the above examples, we observe that how the principal does not remain same always; at the end of every phase, the principal changes. There is a direct relation between the principal and the compound interest or amount.

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