Sequences and Series: AP, GP, AM and GM

Sequences and series

A sequence is a list of numbers in a definite order: a₁, a₂, a₃, …, where aₙ is the nth term (general term). It is finite if it stops and infinite if it goes on forever.

• A sequence can be given by a formula, like aₙ = n² + 1 → 2, 5, 10, 17, …
• Or by a rule linking terms (a recurrence), like a₁ = 1, aₙ = aₙ₋₁ + 2 → 1, 3, 5, 7, …

A series is the sum of the terms: a₁ + a₂ + a₃ + … We write Sₙ for the sum of the first n terms, and use Σ (sigma) as shorthand. A sequence whose terms follow a clear rule is called a progression.

Arithmetic progression and arithmetic mean (AM)

Recall from Class 10: in an AP, each term is the previous one plus a fixed common difference d. aₙ = a + (n − 1)d and Sₙ = n/2[2a + (n − 1)d] = n/2(a + l), where l is the last term.

The arithmetic mean of two numbers a and b is A = (a + b)/2. Then a, A, b form an AP.

Inserting n AMs between a and b: we need n + 2 terms in all, so b = a + (n + 1)d, which gives d = (b − a)/(n + 1). The AMs are a + d, a + 2d, …, a + nd. Example: 3 AMs between 4 and 16 → d = 3 → 7, 10, 13 (step 2 of the 3D).

Geometric progression: general term and sum

A GP is a sequence where each term (after the first) is the previous term times a fixed non-zero number r, the common ratio. So r = a₂/a₁ = a₃/a₂ = …

• General term: aₙ = a rⁿ⁻¹.
• Sum of n terms: Sₙ = a(rⁿ − 1)/(r − 1) or a(1 − rⁿ)/(1 − r), for r ≠ 1. If r = 1, Sₙ = na.

Why the sum formula works: Sₙ = a + ar + … + arⁿ⁻¹. Multiply by r: rSₙ = ar + ar² + … + arⁿ. Subtract: rSₙ − Sₙ = arⁿ − a, so Sₙ = a(rⁿ − 1)/(r − 1). Almost every term cancels.

Tip: three numbers in GP can be taken as a/r, a, ar; then their product is a³.

Infinite GP and its sum

If |r| < 1, then rⁿ gets closer and closer to 0 as n grows (for example (1/2)¹⁰ ≈ 0.001). So Sₙ = a(1 − rⁿ)/(1 − r) gets closer to

S∞ = a/(1 − r).

The sum never goes past this value; it only approaches it (step 4 of the 3D). If |r| ≥ 1, the terms do not shrink and there is no finite infinite sum.

Use: the recurring decimal 0.333… = 3/10 + 3/100 + … = (3/10)/(1 − 1/10) = 1/3.

Geometric mean (GM)

The geometric mean of two positive numbers a and b is G = √(ab). Then a, G, b form a GP.

Inserting n GMs between a and b: b = a rⁿ⁺¹, so r = (b/a)^(1/(n+1)). The GMs are ar, ar², …, arⁿ. Example: 2 GMs between 3 and 81 → r³ = 27, r = 3 → 9, 27.

Relation between AM and GM

For positive numbers a and b, with A = (a + b)/2 and G = √(ab):

A − G = (a + b)/2 − √(ab) = (√a − √b)²/2 ≥ 0, so A ≥ G.

A square can never be negative, so the AM is always at least the GM. They are equal only when √a = √b, i.e. a = b. In the 3D (step 5) the green A bar is taller than the purple G bar.

Use: if the AM and GM of two numbers are known, the numbers are the roots of x² − 2Ax + G² = 0.

Board exam focus

Expect: find a term or the number of terms of a GP (2 marks), sum of n terms or of an infinite GP (2–3 marks), insert AMs or GMs (3 marks), problems using A ≥ G or finding numbers from their AM and GM (3–4 marks), and word problems on growth or bouncing balls (4 marks).

Key formulas and definitions

• AP: aₙ = a + (n − 1)d, Sₙ = n/2[2a + (n − 1)d]
• AM: A = (a + b)/2; n AMs: d = (b − a)/(n + 1)
• GP: aₙ = arⁿ⁻¹, Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1
• Infinite GP, |r| < 1: S∞ = a/(1 − r)
• GM: G = √(ab); n GMs: r = (b/a)^(1/(n+1))
• A ≥ G; A − G = (√a − √b)²/2

Worked examples

Common mistakes

• Using aₙ = arⁿ instead of arⁿ⁻¹. The first term is a = ar⁰.
• Using S∞ = a/(1 − r) when |r| ≥ 1. It is valid only for |r| < 1.
• Finding d for n AMs as (b − a)/n. There are n + 1 gaps, so d = (b − a)/(n + 1).
• Taking the GM of a negative and a positive number. GM is defined here for positive numbers only.