2.12 Compound Interest When the Rate Changes Every Year

Sometimes a question gives a different rate for each year instead of one fixed rate throughout. The fix is simple: multiply by a different factor for each year instead of raising a single factor to a power.

A = P(1+(R_1)/(100))(1+(R_2)/(100))(1+(R_3)/(100))…

This is exactly the “successive percentage change” idea from the Percentage chapter — each year’s growth multiplies onto the previous year’s amount, and the order of multiplication does not matter (multiplication is commutative), so R₁ then R₂ gives the same final amount as R₂ then R₁.

Solved Example 2.12.1: Find the CI on Rs. 10,000 for 2 years, if the rate is 10% for the first year and 20% for the second year.

Solution: A = 10000 × 1.1 × 1.2 = 10000 × 1.32 = Rs. 13,200. CI = 13200 − 10000 = Rs. 3,200.

Solved Example 2.12.2: A sum triples itself in 15 years under simple interest. Find the rate of interest.

Solution: SI = 2P (since amount = 3P means interest earned = 2P). Using T = 100 × (multiple − 1)/R, rearrange: R = 100 × 2/15 = 200/15 = 13.33% p.a. (approx.) This is a direct extension of the doubling-time shortcut (Trick 5) — “triples” means the interest equals 2P instead of 1P, so simply replace 100 with 200 in the numerator.