Hello, and welcome to today's mathematics lesson. We are going to explore compound interest using formulas. In our previous work, we calculated compound interest through repeated simple interest computations, which became quite tedious when dealing with many conversion periods. Today, we discover powerful formulas that make these calculations both easy and fast.
Let us begin with the fundamental formula for annual compounding. When interest is compounded yearly, the amount is given by: A = P(1 + r/100)ⁿ. Here, A represents the final amount, P is the principal, r is the annual rate of interest in percent, and n is the number of years.
The compound interest itself is simply the amount minus the principal: CI = A − P.
Let me walk you through an example. Suppose you invest 7,500 rupees for 2 years at 6 percent compounded annually. Using our formula, we write: A = 7,500 × (1 + 6/100)². This becomes 7,500 times 106 over 100, the quantity squared. Calculating step by step: 106 over 100 is 1.06, and 1.06 squared equals 1.1236. Multiplying 7,500 by 1.1236 gives us 8,427 rupees. Therefore, the compound interest equals 8,427 minus 7,500, which is 927 rupees.
Now, what happens when interest rates change from year to year? We use a modified approach where each year's growth factor multiplies the previous result. The formula becomes: A = P(1 + r₁/100)(1 + r₂/100)(1 + r₃/100).... Each r represents the rate for that specific successive year.
Consider this: 12,000 rupees invested for 3 years with rates of 8 percent, 10 percent, and 15 percent respectively. We calculate: 12,000 times 1.08 times 1.10 times 1.15.
This equals 16,394 rupees and 40 paise.
The compound interest is the difference: 4,394 rupees and 40 paise.
Let us turn to inverse problems, where we work backwards from known information.
First, finding the principal when the amount is known. If a sum grows to 3,630 rupees in 2 years at 10 percent compound interest, what was the original investment? We rearrange our formula: 3,630 equals P times (1 + 10/100)², This is P times 11 over 10, the quantity squared, which equals 121 over 100. Solving for P: we multiply 3,630 by 100 over 121, giving us 3,000 rupees.
Second, finding the rate of interest. When 2,000 rupees becomes 2,315.25 rupees in 3 years, we set up: 2,315.25/2,000 = (1 + r/100)³, The left side simplifies to 1.157625, which equals (21/20)³.
Taking the cube root: 21 over 20 equals 1 plus r over 100. Therefore, r over 100 equals 1 over 20, so r equals 5 percent.
Third, finding the time period. If 2,000 rupees grows to 2,662 rupees at 10 percent compound interest, how many years does it take? We have: 2,662/2,000 = (11/10)ⁿ. This equals 1,331 over 1,000, which is (11/10)³. By comparing exponents, n equals 3 years.
Now we address a special case: when interest is compounded half-yearly. This means interest is calculated twice per year.
The formula adjusts as follows: A = P(1 + r/(2×100))^(n×2) Notice what happens: the annual rate is halved, and the number of years is doubled, giving us the total number of half-year periods.
For example, 4,000 rupees invested for one and a half years at 10 percent compounded half-yearly. Here, n equals 3 over 2, so n times 2 equals 3 half-year periods. The rate per half-year is 10 over 2, which is 5 percent. Thus: A = 4,000 × (1 + 5/100)³, which equals 4,630.50. which calculates to 4,630.50 rupees, making the compound interest 630.50 rupees.
What about fractional years with yearly compounding? When time is not a whole number, we handle complete years first, then apply the half-yearly compound interest formula for the remaining fraction.
Take 10,000 rupees for 2 and a half years at 10 percent compounded yearly. First, calculate for 2 full years: 10,000 × (1.1)², which equals 12,100 rupees. This amount now earns compound interest for the remaining half year at half the annual rate. A = 12,100 × (1 + 10/200), which equals 12,705 rupees.
Let us explore some elegant miscellaneous problems.
Finding principal when the difference between compound and simple interest is known. For 2 years at 5 percent, if this difference is 25 rupees, we proceed as follows. Simple interest equals P times 5 times 2 over 100, which is P over 10. Compound interest equals P(1 + 5/100)² − P, which equals 41 P over 400. Their difference, 41P over 400 minus P over 10, simplifies to P over 400, which equals 25. Therefore, P equals 25 times 400, which is 10,000 rupees.
Another beautiful problem: when amounts after different years are known. If a sum becomes 6,600 in 1 year and 7,986 in 3 years, find the original sum and rate. Dividing the second equation by the first eliminates P: (1 + r/100)² equals 7,986 over 6,600, which equals 121 over 100. Taking square roots, 1 plus r over 100 equals 11 over 10, giving r equals 10 percent. Substituting back, 6,600 equals P times 1 point 1, giving P equals 6,000 rupees.
Finally, we apply our compound interest formula to growth and depreciation problems.
For growth of industries, population increase, or inflation: Production after n years equals initial production times (1 + r/100)ⁿ.
For depreciation, where value decreases: Value after n years equals present value times (1 − r/100)ⁿ.
Consider a machine purchased for 2,50,000 rupees, depreciating 20 percent annually, with scrap value 1,28,000 rupees. We solve: 1,28,000 = 2,50,000 × (4/5)ⁿ. This simplifies to (4/5)ⁿ equals (4/5)³, so n equals 3 years.
For population: if a town has 2 lakh 16,000 people growing 20 percent yearly, after 2 years it becomes 2 lakh 16,000 times 1 point 2 squared, which is 3 lakh 11,040. Two years ago, it was 2 lakh 16,000 divided by 1 point 44, which equals 1 lakh 50,000.
Here is a powerful insight about doubling time. If money doubles in 5 years, then (1 + r/100)⁵ = 2. To become eight times itself, we need (1 + r/100)ⁿ = 8, which is 2 cubed. Substituting, we get (1 + r/100)¹⁵ = 8. so n equals 15 years — exactly three times the doubling period.
Let me summarize the key takeaways from today's lesson.
First, the fundamental formula for annual compounding: A = P(1 + r/100)ⁿ. Second, for varying yearly rates, multiply successive growth factors together. Third, for half-yearly compounding, halve the rate and double the periods. Fourth, inverse problems require careful algebraic manipulation to find principal, rate, or time. Fifth, the same exponential formulas apply to growth and depreciation by using plus or minus in the factor. Sixth, when money doubles in n years, it multiplies by 2 to the power k in k times n years.
Compound interest is one of mathematics' most practical tools, appearing in banking, investments, population studies, and resource management. Master these formulas, and you hold the key to understanding how quantities grow and decay over time. Keep practicing, stay curious, and I look forward to our next mathematical journey together.