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Topics · Arithmetic & Geometric Sequences

Arithmetic and geometric sequences in DSE Maths Paper 2

Sequences start in Section A with number patterns and grow into arithmetic and geometric sequences in Section B. You need the general term and the sum to n terms of each kind, and the sum to infinity of a geometric sequence when it exists.

What this topic covers

  • Number patterns and their general terms
  • General term and sum of an arithmetic sequence
  • General term and sum of a geometric sequence
  • Sum to infinity of a geometric sequence

Two sample questions

Sample 1

The figure shows the first three patterns formed by counters. The 1st pattern is formed by 9 counters, and each pattern after the first has 3 more counters than the one immediately before it. Find the number of counters in the 12 th pattern.

1st2nd3rd
  1. A30
  2. B33
  3. C42
  4. D207
Show the answer and solution

Answer: C

Worked solution

Counts: 9,12,15,…\displaystyle 9, 12, 15, \ldots — an arithmetic sequence with first term 9\displaystyle 9 and common difference 3\displaystyle 3T12=9+(12−1)(3)=9+33=42\displaystyle T_{12} = 9 + (12 - 1)(3) = 9 + 33 = 42

Sample 2

In an arithmetic sequence, the sum of the 3 rd term and the 4 th term is negative 1, and the 8 th term is negative 14. Find the 14 th term of the sequence.

  1. Anegative 53
  2. Bnegative 32
  3. C20
  4. D25
Show the answer and solution

Answer: B

Worked solution

Let a1\displaystyle a_1 be the 1st term and d\displaystyle d the common difference.from a8\displaystyle a_{8}: a1+(8−1)d=−14\displaystyle a_1 + (8-1)d = -14 \quad...(1)from a3+a4\displaystyle a_{3}+a_{4}: 2a1+(3+4−2)d=−1\displaystyle 2a_1 + (3+4-2)d = -1 \quad...(2)(2)−2×\displaystyle - 2\times{}(1):(−9)d=27⇒d=−3\displaystyle : (-9)d = 27 \Rightarrow d = -3a1=−14−(8−1)(−3)=7\displaystyle a_1 = -14 - (8-1)(-3) = 7a14=a1+(14−1)d=−32\displaystyle a_{14} = a_1 + (14-1)d = -32

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Original material by Dsetopia, written from the EDB syllabus learning objectives. Not affiliated with the HKEAA; no exam content is reproduced.