IB Mathematics SL 1.2 Arithmetic Sequences & Series AA SL Paper 1- Exam Style Questions- New Syllabus

(a)
(i) In an arithmetic sequence, the difference between consecutive terms is constant: \(u_2 – u_1 = u_3 – u_2\).
\((3 – 2k) – (k – 5) = (5k + 3) – (3 – 2k)\)
\(3 – 2k – k + 5 = 5k + 3 – 3 + 2k\)
\(8 – 3k = 7k \implies 10k = 8\)
\(k = 0.8\)
(ii) Substituting \(k = 0.8\): \(u_3 = 5(0.8) + 3 = 4 + 3 = 7\).

(b)
(i) For \(k = 12\): \(u_1 = 7\), \(u_2 = -21\), \(u_3 = 63\).
Check the common ratio: \(\frac{-21}{7} = -3\) and \(\frac{63}{-21} = -3\). Since the ratio is constant, the sequence is geometric.
(ii) A geometric series converges only if \(|r| < 1\). Here, \(|r| = 3\). Since \(3 \ge 1\), the sum to infinity does not exist.

(c)
(i) For a geometric sequence, \(u_2^2 = u_1 \times u_3\).
\((3 – 2k)^2 = (k – 5)(5k + 3)\)
\(9 – 12k + 4k^2 = 5k^2 + 3k – 25k – 15\)
\(4k^2 – 12k + 9 = 5k^2 – 22k – 15 \implies k^2 – 10k – 24 = 0\). (Shown)
(ii) Solving \(k^2 – 10k – 24 = 0 \implies (k – 12)(k + 2) = 0\).
The second value is \(k = -2\).
Terms: \(u_1 = -2 – 5 = -7\), \(u_2 = 3 – 2(-2) = 7\), \(u_3 = 5(-2) + 3 = -7\).
(iii) The sequence is \(-7, 7, -7, 7, \dots\). For any even number of terms \(2m\), the terms cancel out in pairs (\(-7 + 7 = 0\)).
\(S_{2m} = 0\).