# Compound Interest as Repeated Simple Interest

We will learn how to calculate compound interest as repeated simple interest.

If the compound interest of any particular year is $z; then the compound interest for the next year on the same sum and at the same rate =$ z + Interest for one year on $z. Thus the compound interest on a principal P for two years = (Simple interest SI on the principal for 1 year) + (simple interest SI' on the new principal (P + SI), that is, the amount at the end of first year, for one year) In the same way, if the amount at compound interest in a particular year is$ z; then the amount for the next year, on the same sum and the same rate = $x + Interest of$ z for one year.

Thus, the compound interest on a principal P for three years = (Simple interest SI on the principal for 1 year) + (simple interest SI' on the new principal (P + SI), that is, the amount at the end of first year, for one year) + (simple interest SI'' on the new principal (P + SI + SI'), that is, the amount at the end of second years, for one year)

This method of calculating compound interest is known as the method of repeated simple interest computation with a growing principal.

In case of simple interest the principal remains the same for the whole period but in case of compound interest the principal changes every year.

Clearly, the compound interest on a principal P for 1 year =simple interest on a principal for 1 year, when the interest is calculated yearly.

The compound interest on a principal for 2 years > the simple interest on the same principal for 2 years.

Remember, if the principal = P, amount at the end of the period = A and compound interest = CI,      CI = A - P

Solved examples on Compound Interest as Repeated Simple Interest:

1. Find the compound interest on $14000 at the rate of interest 5% per annum. Solution: Interest for the first year = $$\frac{14000 × 5 × 1}{100}$$ =$700

Amount at the end of first year = $14000 +$700

= $14700 Principal for the second year =$14700

Interest for the second year = $$\frac{14700 × 5 × 1}{100}$$

= $735 Amount at the end of second year =$14700 + $735 =$15435

Therefore, compound interest = A – P

= final amount – original principal

= $15435 -$14000

= $1435 2. Find the compound interest on$30000 for 3 years at the rate of interest 4% per annum.

Solution:

Interest for the first year = $$\frac{30000 × 4 × 1}{100}$$

= $1200 Amount at the end of first year =$30000 + $1200 =$31200

Principal for the second year = $31200 Interest for the second year = $$\frac{31200 × 4 × 1}{100}$$ =$1248

Amount at the end of second year = $31200 +$1248

= $32448 Principal for the third year =$32448

Interest for the third year = $$\frac{32448 × 4 × 1}{100}$$

= $1297.92 Amount at the end of third year =$32448 + $1297.92 =$33745.92

Therefore, compound interest = A – P

= final amount – original principal

= $33745.92 -$30000

= $3745.92 3. Calculate the amount and compound interest on$10000 for 3 years at 9% p.a.

Solution:

Interest for the first year = $$\frac{10000 × 9 × 1}{100}$$

= $900 Amount at the end of first year =$10000 + $900 =$10900

Principal for the second year = $10900 Interest for the second year = $$\frac{10900 × 9 × 1}{100}$$ =$981

Amount at the end of second year = $10900 +$981

= $11881 Principal for the third year =$11881

Interest for the third year = $$\frac{11881 × 9 × 1}{100}$$

= $1069.29 Amount at the end of third year =$11881 + $1069.29 =$12950.29

Therefore, the required amount = $12950.29 Therefore, compound interest = A – P = final amount – original principal =$12950.29 - $10000 =$2950.29