IB Mathematics SL 1.2 Arithmetic sequences and series AI SL Paper 2 - Exam Style Questions - New Syllabus

(a)
The distance each day is found by multiplying the previous day’s distance by a constant factor (1.05, representing a 5% increase). Therefore, the distances form a geometric sequence.
Common ratio: \( r = 1.05 \).
\(\boxed{\text{Geometric sequence because each term is multiplied by a constant. } r = 1.05}\)

(b)
First term \(a = 2\), common ratio \(r = 1.05\), number of terms \(n = 15\).
Sum of geometric series: \( S_n = \frac{a(r^n – 1)}{r – 1} \).
\( S_{15} = \frac{2(1.05^{15} – 1)}{1.05 – 1} \approx 43.2 \text{ km} \).
\(\boxed{43.2 \text{ km}}\)

(c)
Total runs in 12 weeks: \(3 \times 12 = 36\) runs.
\(u_n = ar^{n-1}\), so \(u_{36} = 2 \times 1.05^{35} \approx 11.0 \text{ km}\).
Race distance = \(5 \text{ km}\), twice race distance = \(10 \text{ km}\).
Since \(11.0 > 10\), Alex’s final run meets the recommendation.
\(\boxed{u_{36} \approx 11.0 \text{ km} > 10 \text{ km}}\)

(d)(i)
In week 1: Each swim = \(80 \times 2 = 160 \text{ m}\) (out and back).
Two swims per week: \(u_1 = 160 \times 2 = 320 \text{ m}\).
\(\boxed{320 \text{ m}}\)

(d)(ii)
Arithmetic sequence: \(u_n = u_1 + (n-1)d\).
Given \(u_4 = 704\):
\(704 = 320 + (4-1)d \implies 704 = 320 + 3d \implies 3d = 384 \implies d = 128\).
Weekly increase = \(128 \text{ m}\).
\(\boxed{128 \text{ m}}\)

(d)(iii)
In week 2: Total distance \(u_2 = u_1 + d = 320 + 128 = 448 \text{ m}\).
This is for two swims, each swim = \(448 \div 2 = 224 \text{ m}\).
Each swim is out and back, so one-way distance = \(224 \div 2 = 112 \text{ m}\).
\(\boxed{112 \text{ m}}\)

(e)
For week 12: \(u_{12} = u_1 + (12-1)d = 320 + 11 \times 128 = 320 + 1408 = 1728 \text{ m}\).
This is total for the week. Each swim = \(1728 \div 2 = 864 \text{ m}\).
\(\boxed{864 \text{ m}}\)