9.2 Sequences and Series

Sequences without a hence

Sequences and series sit under MAT algebraic processes on the public Test Content list: recognise and use patterns, including sequences and series. The 2015 MAT teachers' booklet also names sigma notation. This is still the MAT afternoon paper, not QL. QL may ask you to continue a table of campus counts. MAT asks you to name the nth term, a partial sum, or a sigma rewrite, and it will not scaffold you with write the first three terms. Hence find S_10.

There is no calculator. The official 2023 Intake National Report describes MAT as 60 items; the shorter current Test Content page does not restate that count, so current session instructions remain controlling. Independent OpenExamPrep sequence practice uses simple integers so every line can be checked by hand. It does not copy confidential NBTP items.

A sequence is an ordered list T_1, T_2, T_3, …. A series is a sum of terms of a sequence. The same numbers can be either, depending on whether the stem wants the 8th term or the sum of eight terms. Mixing those two jobs is the cheapest way to pick a distractor that was written on purpose.

Arithmetic sequences

An arithmetic sequence has a constant difference d between consecutive terms.

• General term: T_n = a + (n − 1)d, where a is the first term
• Sum of n terms: S_n = n/2 × [2a + (n − 1)d], or S_n = n/2 × (first + last)

Worked example: eighth term. First term a = 5, common difference d = 3.

T_8 = 5 + (8 − 1) × 3 = 5 + 21 = 26.

Traps: 5 + 8 × 3 = 29 uses n instead of n − 1. T_2 = 8 is only the second term. 5 × 3 × 8 treats the sequence as a product.

Worked example: sum of the first eight terms of the same sequence. Last term T_8 = 26. Then S_8 = 8/2 × (5 + 26) = 4 × 31 = 124.

Formula check: S_8 = 8/2 × [2 × 5 + 7 × 3] = 4 × (10 + 21) = 4 × 31 = 124. Adding 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 also gives 124. On MAT you will not have time to add twenty terms; the formula is the method.

Worked example: recover d, then T_10. Terms 11, 17, 23, 29, … Difference 17 − 11 = 6. Check 23 − 17 = 6. T_n = 11 + (n − 1) × 6 = 6n + 5. T_10 = 11 + 9 × 6 = 11 + 54 = 65.

Worked example: find n. An arithmetic sequence starts at 4 with d = 5 and T_n = 64. Then 4 + (n − 1) × 5 = 64, so (n − 1) × 5 = 60, n − 1 = 12, n = 13. Using n × 5 = 60 and quoting n = 12 drops the first term.

Geometric sequences

A geometric sequence has a constant ratio r (r ≠ 0).

• General term: T_n = a r^{n−1}
• Finite sum, r ≠ 1: S_n = a(r^n − 1)/(r − 1)
• If |r| < 1, the infinite sum S_∞ = a/(1 − r) exists

Worked example: fifth term and sum. Sequence 3, 6, 12, 24, … Ratio r = 6/3 = 2.

T_5 = 3 × 2^{4} = 3 × 16 = 48. S_5 = 3(2^5 − 1)/(2 − 1) = 3(32 − 1) = 3 × 31 = 93. Line check: 3 + 6 + 12 + 24 + 48 = 93.

A writer who reports 48 when the stem asked for the sum has answered T_5. A writer who computes 3 × 2^5 = 96 has used n as the exponent on the last power and then multiplied by a instead of summing. 96 sits 3 away from 93, which is why it is a tempting neighbour. 5 × 3 = 15 is an arithmetic-looking product of n and a.

Worked example: fractional ratio. 80, 40, 20, 10, … Here r = 1/2, a = 80.

T_6 = 80 × (1/2)^5 = 80/32 = 2.5. S_4 = 80(1 − (1/2)^4)/(1 − 1/2) = 80(1 − 1/16)/(1/2) = 80 × (15/16) × 2 = 10 × 15 = 150. Check: 80 + 40 + 20 + 10 = 150.

Infinite case, same first term. |r| = 1/2 < 1, so S_∞ = 80 / (1 − 1/2) = 80 / (1/2) = 160. The finite S_4 = 150 already sits close to 160; the missing tail is 10 + 5 + 2.5 + … = 10. Using the infinite formula when |r| ≥ 1 is undefined as a real sum: 3, 6, 12, … does not have a finite S_∞.

Sigma notation

Sigma notation is a compact series. Σ from k = 1 to n of u_k means add u_k from k = 1 through k = n.

Identities you should compute, not decorate:

• Σ_{k=1}^{n} c = n c
• Σ_{k=1}^{n} k = n(n + 1)/2
• Σ_{k=1}^{n} (ak + b) splits as a Σ k + b Σ 1

Worked example: linear sum. Σ_{k=1}^{4} (2k + 1).

Terms: k = 1 → 3; k = 2 → 5; k = 3 → 7; k = 4 → 9. Sum: 3 + 5 + 7 + 9 = 24.

Formula: 2 Σ k + Σ 1 = 2 × (4 × 5 / 2) + 4 × 1 = 2 × 10 + 4 = 24. A writer who computes 2(1 + 2 + 3 + 4) = 20 dropped the +1 on each term. A writer who reports 9 quoted the last term. A writer who reports 2 × 4 + 1 = 9 substituted n into the general term and stopped.

Worked example: geometric sigma. Σ_{k=0}^{3} 5 × 2^k. This is a + ar + ar^2 + ar^3 with a = 5, r = 2, four terms because k runs from 0 through 3.

S = 5(2^4 − 1)/(2 − 1) = 5 × 15 = 75. Terms: 5 + 10 + 20 + 40 = 75.

Index trap: Σ_{k=1}^{3} 5 × 2^k would be 10 + 20 + 40 = 70, missing the k = 0 term. Read the limits.

Worked example: sigma as an arithmetic sum. Σ_{k=1}^{10} (3k − 1) is an arithmetic series with first term 2, last term 29, ten terms. S_10 = 10/2 × (2 + 29) = 5 × 31 = 155. Split: 3 × (10 × 11 / 2) − 10 = 3 × 55 − 10 = 165 − 10 = 155.

Which pattern is it?

Worked example: not everything is AP or GP. 2, 6, 12, 20, 30, … First differences: 4, 6, 8, 10, which themselves form an arithmetic sequence. Second differences are constant, so this is a quadratic pattern: T_n = n(n + 1). T_6 = 6 × 7 = 42. Forcing an arithmetic d = 4 from the first step already fails at the third term (that guess would predict 10, not 12).

Sitting the item

Cover the options. Decide arithmetic, geometric, or neither. Write T_n or S_n with n − 1 in the right place. Expand a short sigma by hand if n is small. Then look for your number. MAT will not say hence. QL will not save you with a printed series formula.

Traps

• n versus n − 1 in T_n
• Sum versus term
• Arithmetic formula used on a geometric sequence
• Sigma limits off by one
• Infinite-sum formula used when |r| ≥ 1
• Calculator habit: 2^{10} is still 1024 by hand, because 2^5 = 32 and 32^2 = 1024

Independent OpenExamPrep MAT series teaching is the formula you can defend on paper, not a list of school hence sub-questions.