Geometric progression and its nth term
A ball that rebounds to half its height each time, or an amount that doubles every year: each step multiplies by the same number.
A sequence is a geometric progression (G.P.) if each term after the first is obtained by multiplying the previous term by a fixed number r, the common ratio. With first term a: aₙ = a rⁿ⁻¹.
Formula: aₙ = a rⁿ⁻¹, r = a₂/a₁
Rule 1: Common ratio — Divide any term by the previous term.
- Easy: 2, 6, 18, 54: r = 3.
- Medium: 80, 40, 20, 10: r = 1/2.
- Exam-type: 3, −6, 12, −24: r = −2.
Rule 2: nth term — Put the values in a rⁿ⁻¹.
- Easy: a = 2, r = 3: a₅ = 2 × 3⁴ = 162.
- Medium: a = 64, r = 1/2: a₇ = 64 × (1/2)⁶ = 1.
- Exam-type: a = 5, r = 2: which term is 640? 5 × 2ⁿ⁻¹ = 640 gives 2ⁿ⁻¹ = 128, n = 8.
To prove three numbers a, b, c are in G.P., check that b² = ac.
Using r = a₂ − a₁ (that is the common difference of an A.P.).
G.P.: Go by Product, Add for A.P.
Spot the difference
| A.P. | G.P. |
|---|---|
| Add the same number | Multiply by the same number |
| d = a₂ − a₁ | r = a₂ / a₁ |
Practice questions
- 1Find the common ratio of 4, 12, 36, 108.
- 2Find the 6th term of 3, 6, 12, ….
- 3Find the 5th term of 81, 27, 9, ….
- 4Which term of 2, 4, 8, … is 512?
- 5The 3rd term of a G.P. is 12 and the 6th term is 96. Find a and r.
- 6Find the 8th term of 1, −2, 4, ….
- 7Are 2, 6, 18 and 4, 6, 9 each in G.P.?
- 8Find the positive x for which 2, x, 18 are in G.P.
Show answers (try them yourself first)
- 13
- 296
- 31
- 42ⁿ = 512, n = 9
- 5r³ = 8, r = 2, a = 3
- 6-128
- 72, 6, 18: r = 3 yes. 4, 6, 9: r = 3/2 yes.
- 8x² = 2 × 18 = 36, so x = 6
Sum of a G.P.
If you save ₹1 on day 1, ₹2 on day 2, ₹4 on day 3 and keep doubling, how much have you saved in 10 days? A formula adds a G.P. quickly.
The sum of the first n terms of a G.P. with first term a and common ratio r ≠ 1 is Sₙ = a(rⁿ − 1)/(r − 1) (or a(1 − rⁿ)/(1 − r)). If r = 1, Sₙ = na.
Formula: Sₙ = a (rⁿ − 1)/(r − 1)
Rule 1: Sum of n terms — Find a, r and n, then use the formula.
- Easy: 1 + 2 + 4 + … (10 terms): S = (2¹⁰ − 1)/(2 − 1) = 1023.
- Medium: 3 + 6 + 12 + … (6 terms): S = 3(2⁶ − 1) = 189.
- Exam-type: 81 + 27 + 9 + … (5 terms): S = 81(1 − (1/3)⁵)/(1 − 1/3) = 121.
Rule 2: Find n or a missing value — Rearrange the formula.
- Easy: 1 + 3 + 9 + … + 243 has n = 6 terms (3⁵ = 243).
- Medium: Sum = 3(3ⁿ − 1)/2 = 363 gives 3ⁿ − 1 = 242, n = 5.
- Exam-type: If a = 2, r = 3 and Sₙ = 80, then 3ⁿ = 81, n = 4.
Check the formula with 2 terms: a + ar = a(r² − 1)/(r − 1) = a(r + 1).
Using the formula with r = 1 (it divides by zero).
First × (power − 1) ÷ (ratio − 1).
Practice questions
- 1Find the sum of 1 + 2 + 4 + 8 + 16.
- 2Find the sum of 2 + 6 + 18 + 54.
- 3Find the sum of the first 5 terms of 16, 8, 4, ….
- 4Find the sum of the first 7 terms of 1, −2, 4, ….
- 5How many terms of 1, 3, 9, … give a sum of 364?
- 6Find 4 + 12 + 36 + … + 2916. (find n first)
- 7A tree is 2 m tall. Each year the new growth is 3/4 of last year's (first year growth 1 m). Find the total height after 3 years of growth.
- 8If the sum of the first 3 terms of a G.P. is 7 and r = 2, find the first term.
Show answers (try them yourself first)
- 131
- 280
- 331
- 443
- 5(3ⁿ − 1)/2 = 364, 3ⁿ = 729, n = 6
- 64·3ⁿ⁻¹ = 2916 gives 3ⁿ⁻¹ = 729, n = 7; S = 4(3⁷ − 1)/2 = 4372
- 72 + 1 + 0.75 + 0.5625 = 4.3125
- 8a(2³ − 1) = 7 gives a = 1
Geometric mean
Two numbers on a ruler: the number that makes the three steps grow by the same ratio sits between them. That is the geometric mean.
The geometric mean (G.M.) of two positive numbers a and b is √(ab); it is the middle term of the G.P. a, G, b. To insert n geometric means between a and b, the common ratio is r = (b/a)^(1/(n+1)).
Formula: G = √(ab)
Rule 1: G.M. of two numbers — Multiply and take the square root.
- Easy: G.M. of 4 and 9 = √36 = 6.
- Medium: G.M. of 2 and 8 = √16 = 4.
- Exam-type: G.M. of 3 and 27 = √81 = 9.
Rule 2: Insert means — Find r first.
- Easy: Insert 2 G.M.s between 3 and 81: r³ = 27, r = 3, so 3, 9, 27, 81.
- Medium: Insert 1 G.M. between 5 and 20: √100 = 10, so 5, 10, 20.
- Exam-type: Insert 3 G.M.s between 1 and 16: r⁴ = 16, r = 2, so 1, 2, 4, 8, 16.
The G.M. is always less than or equal to the A.M. (next topic).
Taking the average instead of √(ab).
Multiply, then root.
Practice questions
- 1Find the G.M. of 4 and 16.
- 2Find the G.M. of 2 and 50.
- 3Insert 2 G.M.s between 2 and 54.
- 4Insert 3 G.M.s between 2 and 32.
- 5Find the G.M. of 1/2 and 8.
- 6Insert 4 G.M.s between 1 and 243.
- 7The A.M. of two numbers is 10 and the G.M. is 8. Find the numbers.
- 8Show that 6 is the G.M. of 4 and 9 and of 3 and 12.
Show answers (try them yourself first)
- 18
- 210
- 3r³ = 27, r = 3: 2, 6, 18, 54
- 4r⁴ = 16, r = 2: 2, 4, 8, 16, 32
- 52
- 6r⁵ = 243, r = 3: 1, 3, 9, 27, 81, 243
- 7a + b = 20, ab = 64: a, b = 16, 4
- 84 × 9 = 36 and 3 × 12 = 36, so G.M. = 6 in both
Relation between A.M. and G.M.
Of all rectangles with the same perimeter, the square has the largest area. This is a picture of A.M. ≥ G.M.
The arithmetic mean (A.M.) of a and b is (a + b)/2. For positive numbers, A.M. ≥ G.M.: (a + b)/2 ≥ √(ab), with equality only when a = b.
Formula: (a + b)/2 ≥ √(ab)
Rule 1: Compare A.M. and G.M. — Compute both and compare.
- Easy: a = 4, b = 9: A.M. = 6.5 ≥ G.M. = 6.
- Medium: a = 5, b = 5: A.M. = G.M. = 5.
- Exam-type: a = 1, b = 16: A.M. = 8.5 ≥ G.M. = 4.
Rule 2: Use the inequality — Maximise or minimise using the sum or product.
- Easy: If x + y = 10 (positive), xy ≤ 25, largest when x = y = 5.
- Medium: If xy = 36 (positive), x + y ≥ 12.
- Exam-type: For x > 0: x + 1/x ≥ 2.
A.M. − G.M. = (√a − √b)²/2 ≥ 0, which proves the result.
Using A.M. ≥ G.M. for negative numbers.
Arithmetic is Always at least Geometric.
Practice questions
- 1Find A.M. and G.M. of 2 and 8.
- 2Which is larger for 3 and 12: A.M. or G.M.?
- 3If x + y = 20 with x, y > 0, find the maximum of xy.
- 4If xy = 49, x, y > 0, find the minimum of x + y.
- 5Show that A.M. = G.M. when a = b.
- 6The A.M. and G.M. of two numbers are 25 and 24. Find the numbers.
- 7Find the minimum value of x + 4/x for x > 0.
- 8A rectangle has perimeter 40 m. Find the maximum area.
Show answers (try them yourself first)
- 1A.M. = 5, G.M. = 4
- 2A.M. = 7.5, G.M. = 6, so A.M.
- 3100 (x = y = 10)
- 414 (x = y = 7)
- 5(a + a)/2 = a and √(a·a) = a
- 6a + b = 50, ab = 576: a, b = 32, 18
- 7x + 4/x ≥ 2√4 = 4, at x = 2
- 8l + w = 20, lw ≤ 100 m², when l = w = 10 m