IB Mathematics AHL 1.11 The sum of infinite geometric sequences AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Find the third term of the infinite geometric sequence:
Method 1:
Let the first term be \( a \), common ratio \( r \) (\( |r| < 1 \)).
Sum of infinite series: \( \frac{a}{1 – r} = 9 \). [1]
First term is 4 more than second term (\( ar \)): \( a = ar + 4 \), so \( a(1 – r) = 4 \). [1]
From sum: \( a = 9(1 – r) \). Substitute: \( 9(1 – r)(1 – r) = 4 \), so \( 9(1 – r)^2 = 4 \), \( (1 – r)^2 = \frac{4}{9} \), \( 1 – r = \pm \frac{2}{3} \). [1]
Solve: \( r = \frac{1}{3} \) (since \( |r| < 1 \)). Then \( a = 9 \left(1 – \frac{1}{3}\right) = 6 \). Third term: \( ar^2 = 6 \cdot \left(\frac{1}{3}\right)^2 = \frac{6}{9} = \frac{2}{3} \). [1]
Thus: Third term = \( \frac{2}{3} \). [4]
Method 2:
Sum: \( \frac{a}{1 – r} = 9 \). [1]
First term condition: \( a = ar + 4 \), so \( r = \frac{a – 4}{a} \). [1]
Substitute: \( \frac{a}{1 – \frac{a – 4}{a}} = 9 \), simplify: \( \frac{a}{\frac{4}{a}} = 9 \), so \( a^2 = 36 \), \( a = \pm 6 \). For \( a = 6 \), \( r = \frac{6 – 4}{6} = \frac{1}{3} \) (valid since \( |r| < 1 \)). [1]
Third term: \( ar^2 = 6 \cdot \left(\frac{1}{3}\right)^2 = \frac{2}{3} \). [1]
Thus: Third term = \( \frac{2}{3} \). [4]
Total: [4 marks]
▶️ Answer/Explanation
a) Express the general term \( u_n \):
The sequence is geometric with first term \( a = 10 \) and common ratio \( r = \frac{5}{10} = 0.5 \). The general term is \( u_n = a \times r^{n-1} = 10 \times 0.5^{n-1} \). [1]
Thus: \( u_n = 10 \times 0.5^{n-1} \) (or \( 20 \times 0.5^n \)). [1]
b) Find the first term smaller than \( 10^{-3} = 0.001 \):
Set \( u_n = 10 \times 0.5^{n-1} < 0.001 \). [1]
Solve: \( 10 \times 0.5^{n-1} < 0.001 \), so \( 0.5^{n-1} < 0.0001 \). Since \( 0.5 = 2^{-1} \), \( 0.5^{n-1} = 2^{-(n-1)} \), so \( 2^{-(n-1)} < 10^{-4} \). [1]
Take logs: \( -(n-1) \times \log 2 < -4 \), \( n-1 > \frac{4}{\log 2} \approx 13.288 \), so \( n > 14.288 \). Thus, \( n = 15 \). Check: \( u_{15} = 10 \times 0.5^{14} \approx 0.000610 < 0.001 \), \( u_{14} = 0.001221 > 0.001 \). [1]
Thus: The first term is \( u_{15} = 0.000610 \). [3]
c) Find the sum of the first 20 terms to 6 decimal places:
Sum of a geometric series: \( S_n = a \times \frac{1 – r^n}{1 – r} \). For \( n = 20 \), \( a = 10 \), \( r = 0.5 \): \( S_{20} = 10 \times \frac{1 – 0.5^{20}}{1 – 0.5} \). [1]
Calculate: \( 0.5^{20} \approx 9.53674 \times 10^{-7} \), so \( 1 – 0.5^{20} \approx 0.999999046 \), \( S_{20} = 10 \times \frac{0.999999046}{0.5} \approx 19.999981 \). [1]
Thus: \( S_{20} = 19.999981 \) (to 6 decimal places). [2]
d) Find the sum of the infinite series:
For an infinite geometric series (\( |r| < 1 \)): \( S_\infty = \frac{a}{1 – r} \). With \( a = 10 \), \( r = 0.5 \): \( S_\infty = \frac{10}{1 – 0.5} = \frac{10}{0.5} = 20 \). [1]
Thus: \( S_\infty = 20 \). [2]
Total: [8 marks]