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Number and Algebra - Applications of G.S. – Percentage growth

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Geometric Sequence (G.S.) is a sequence where each term is found by multiplying the previous term by a constant called the common ratio rr. This models percentage growth or decay perfectly.

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For percentage growth of p%p\%, the common ratio is r=1+p100r = 1 + \frac{p}{100}. For example, a 5%5\% increase corresponds to r=1.05r = 1.05.

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For percentage decay or depreciation of p%p\%, the common ratio is r=1−p100r = 1 - \frac{p}{100}. For example, a 12%12\% decrease corresponds to r=0.88r = 0.88.

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In financial applications like compound interest, the value of an investment after nn periods is modeled by the nn-th term of a G.S. If u1u_1 is the initial amount (at t=0t=0), then the amount after nn time periods is un+1=u1×rnu_{n+1} = u_1 \times r^n.

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Compound interest can be calculated more than once a year. If the annual rate is r%r\% and it is compounded kk times per year for nn years, the multiplier for each period is (1+r100k)(1 + \frac{r}{100k}) and the total number of periods is knkn.

📐Formulae

un=u1rn−1u_n = u_1 r^{n-1}

r=1±p100r = 1 \pm \frac{p}{100}

FV=PV×(1+r100k)knFV = PV \times \left(1 + \frac{r}{100k}\right)^{kn}

Vn=V0×(1−r)nV_n = V_0 \times (1 - r)^n

💡Examples

Problem 1:

A population of bacteria starts at 50005000 and increases by 8%8\% every hour. Calculate the population after 66 hours.

Solution:

Let u1=5000u_1 = 5000 be the initial population at t=0t = 0. The growth rate is p=8%p = 8\%, so the common ratio is r=1+8100=1.08r = 1 + \frac{8}{100} = 1.08. To find the population after 66 hours, we need the term after 66 growth steps, which is u7u_7. u7=u1×r6u_7 = u_1 \times r^6 u7=5000×(1.08)6u_7 = 5000 \times (1.08)^6 u7≈5000×1.58687u_7 \approx 5000 \times 1.58687 u7≈7934.37u_7 \approx 7934.37 The population after 66 hours is approximately 79347934.

Explanation:

Since the population increases every hour, it forms a geometric sequence. After nn hours, we have multiplied the initial value by the ratio nn times.

Problem 2:

A car is purchased for 2500025000. It depreciates in value by 15%15\% each year. Find the value of the car after 44 years.

Solution:

The initial value V0=25000V_0 = 25000. The rate of decay is p=15%p = 15\%, so r=1−0.15=0.85r = 1 - 0.15 = 0.85. The value after n=4n = 4 years is: V4=25000×(0.85)4V_4 = 25000 \times (0.85)^4 Calculation: (0.85)4=0.52200625(0.85)^4 = 0.52200625 V4=25000×0.52200625=13050.15625V_4 = 25000 \times 0.52200625 = 13050.15625 The value after 44 years is 13050.1613050.16.

Explanation:

Depreciation is modeled by a geometric sequence where the common ratio is less than 11.

Problem 3:

An investment of 80008000 earns 4%4\% interest per annum, compounded quarterly. Find the total value of the investment after 33 years.

Solution:

Initial Principal PV=8000PV = 8000. Annual rate r=4r = 4. Compounding periods per year k=4k = 4 (quarterly). Total years n=3n = 3. The periodic interest rate is 4%4=1%\frac{4\%}{4} = 1\%, so the multiplier is 1.011.01. The total number of compounding periods is 4×3=124 \times 3 = 12. FV=8000×(1+4100×4)4×3FV = 8000 \times \left(1 + \frac{4}{100 \times 4}\right)^{4 \times 3} FV=8000×(1.01)12FV = 8000 \times (1.01)^{12} FV≈8000×1.126825FV \approx 8000 \times 1.126825 FV≈9014.60FV \approx 9014.60 Total value is 9014.609014.60.

Explanation:

When interest is compounded quarterly, we divide the annual rate by 44 and multiply the number of years by 44 to find the total number of growth steps.