IBDP Maths SL 1.2 Arithmetic Sequences & Series AA HL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AA : HL Paper -1 : All Topics

Question

Consider a sequence \( \{u_n\} \) where the first three terms are defined as: \[ u_1 = k – 5, \quad u_2 = 3 – 2k, \quad \text{and} \quad u_3 = 5k + 3, \] where \( k \) is a real constant.
(a) Assume the sequence \( \{u_n\} \) is arithmetic.
 (i) Determine the value of \( k \).
 (ii) Hence, find the value of the third term, \( u_3 \).
(b) Now consider the case where \( k = 12 \).
 (i) Show that \( u_1, u_2, \) and \( u_3 \) form the start of a geometric sequence.
 (ii) For this geometric sequence, explain why the sum to infinity, \( S_\infty \), does not exist.
(c) The sequence is also geometric for a different value of \( k \).
 (i) Show that this second value of \( k \) satisfies the equation \( k^2 – 10k – 24 = 0 \).
 (ii) Calculate the first three terms of the sequence for this second value of \( k \).
 (iii) Write down the sum of the first \( 2m \) terms, \( S_{2m} \), for this sequence.

Syllabus Topic Codes (IB Mathematics AA HL):

• SL 1.2: Arithmetic sequences and series; use of the formulae for the \( n \)th term and the sum — part (a)
• SL 1.3: Geometric sequences and series; sum of a finite geometric series; sum to infinity — parts (b), (c)
▶️ Answer/Explanation

(a)
For an arithmetic sequence, the common difference is constant: \( u_2 – u_1 = u_3 – u_2 \).
\( (3 – 2k) – (k – 5) = (5k + 3) – (3 – 2k) \)
\( 8 – 3k = 7k \implies 10k = 8 \implies k = \frac{4}{5} \).
✅ (i) Answer: \( \boxed{k = 0.8} \)
Substitute \( k = 0.8 \) into \( u_3 = 5(0.8) + 3 = 4 + 3 = 7 \).
✅ (ii) Answer: \( \boxed{u_3 = 7} \)

(b)
For \( k = 12 \): \( u_1 = 7, u_2 = -21, u_3 = 63 \).
Checking ratios: \( \frac{-21}{7} = -3 \) and \( \frac{63}{-21} = -3 \).
As the common ratio \( r = -3 \) is constant, it is a geometric sequence.
The sum to infinity exists only if \( |r| < 1 \). Since \( |-3| = 3 > 1 \), the series diverges.
✅ (ii) Answer: \( \boxed{|r| \ge 1} \)

(c)
For a geometric sequence: \( u_2^2 = u_1 \cdot u_3 \).
\( (3 – 2k)^2 = (k – 5)(5k + 3) \)
\( 9 – 12k + 4k^2 = 5k^2 + 3k – 25k – 15 \)
\( 9 – 12k + 4k^2 = 5k^2 – 22k – 15 \)
\( 0 = k^2 – 10k – 24 \).
(i) Shown: \( k^2 – 10k – 24 = 0 \).
Factorizing: \( (k – 12)(k + 2) = 0 \). The second value is \( k = -2 \).
Substituting \( k = -2 \):
\( u_1 = -2 – 5 = -7 \)
\( u_2 = 3 – 2(-2) = 7 \)
\( u_3 = 5(-2) + 3 = -7 \)
✅ (ii) Answer: \( \boxed{-7, 7, -7} \)
For this sequence, \( r = -1 \). The sum \( S_{2m} \) of an even number of terms in an oscillating sequence like \( -7, 7, -7, 7, \dots \) always results in zero as each pair sums to zero.
✅ (iii) Answer: \( \boxed{S_{2m} = 0} \)

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