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CIE A-Level Maths Study Notes

1.6.4 Geometric Progressions

Geometric Progression (GP) is a sequence where each term is multiplied by a constant value, known as the common ratio. This section explores formulas for the nth term and the sum of the first n terms of a GP, the defining property of GPs, and examples of finding terms and sums.

Key Concepts

  • Definition: A sequence where each term after the first is obtained by multiplying the previous term by a constant value, known as the common ratio. Example: 2, 4, 8, 16, 32...
  • Nth Term Formula: un=a×rn1u_n = a \times r^{n-1}
    • unu_n: nth term of the sequence
    • aa: First term of the sequence
    • nn: Number of terms
    • rr: Common Ratio
  • Sum of First n Terms: Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r}
    • SnS_n: Sum of the first n terms
  • Sum to Infinity for ( |r| < 1 ): S=a1rS_{\infty} = \frac{a}{1 - r}
Geometric progression

Image courtesy of Cuemath

Examples

Example 1: Finding a Specific Term in a Geometric Progression

Question: Calculate the profit for the year 2008 for a company with an initial profit of £250,000 in 2000, increasing annually by 5%.

Solution:

Profit for 2008 (9th term) in the GP:

u9=250000×1.0591=£369,363.86 (approximately)u_9 = 250000 \times 1.05^{9-1} = £369,363.86 \text{ (approximately)}

Example 2: Finding the Sum of First n Terms in a Geometric Progression

Question: Determine the total profit from 2000 to 2009 under a 5% annual increase.

Solution:

Total profit for 10 years:

S10=250000(11.0510)11.05=£3,144,473.13 (approximately)S_{10} = \frac{250000(1 - 1.05^{10})}{1 - 1.05} = £3,144,473.13 \text{ (approximately)}

Example 3: Comparing Geometric with Arithmetic Progression

Question: For an equal total profit in 10 years under a constant annual increase (Plan B), find the value of DD.

Solution:

Equating the sum under Plan A to that of an arithmetic progression:

3,144,473.13=12×10×[2×250000+(101)×D]3,144,473.13 = \frac{1}{2} \times 10 \times [2 \times 250000 + (10-1) \times D]D=£14,321.63 (approximately)D = £14,321.63 \text{ (approximately)}

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