IBDP Maths AA SL 1.8 The sum of infinite geometric sequences AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)(i) Geometric series: ratio between consecutive terms is constant.
\( \frac{p \ln x}{\ln x} = \frac{\frac{1}{3} \ln x}{p \ln x} \implies \frac{p}{1} = \frac{\frac{1}{3}}{p} \implies p^2 = \frac{1}{3} \implies p = \pm \frac{1}{\sqrt{3}} \).
(a)(ii) For convergence, \( |r| < 1 \). Here, \( r = p = \pm \frac{1}{\sqrt{3}} \), so \( |p| = \frac{1}{\sqrt{3}} < 1 \).
Thus, the geometric series converges.
(a)(iii) Given \( p = \frac{1}{\sqrt{3}} \), sum: \( S_\infty = \frac{\ln x}{1 – \frac{1}{\sqrt{3}}} = 3 + \sqrt{3} \).
Simplify: \( \ln x \cdot \frac{\sqrt{3}}{\sqrt{3} – 1} = 3 + \sqrt{3} \).
Rationalize: \( \ln x \cdot \frac{\sqrt{3} (\sqrt{3} + 1)}{2} = 3 + \sqrt{3} \implies \ln x \cdot \frac{3 + \sqrt{3}}{2} = 3 + \sqrt{3} \implies \ln x = 2 \).
Thus, \( x = e^2 \).
(b)(i) Arithmetic series: \( p \ln x – \ln x = \frac{1}{3} \ln x – p \ln x \).
Simplify: \( p – 1 = \frac{1}{3} – p \implies 2p = \frac{4}{3} \implies p = \frac{2}{3} \).
(b)(ii) Common difference: \( d = p \ln x – \ln x = \frac{2}{3} \ln x – \ln x = -\frac{1}{3} \ln x \).
Thus, \( d = -\frac{1}{3} \ln x \).
(b)(iii) Sum: \( S_n = \frac{n}{2} \left[ 2 \ln x + (n – 1) \left( -\frac{1}{3} \ln x \right) \right] = \ln \left( \frac{1}{x^3} \right) = -3 \ln x \).
Simplify: \( \frac{n}{2} \left[ 2 – \frac{n – 1}{3} \right] \ln x = -3 \ln x \implies \frac{n}{2} \left( \frac{7 – n}{3} \right) = -3 \).
Solve: \( n (7 – n) = -18 \implies n^2 – 7n + 18 = 0 \implies (n – 9)(n + 2) = 0 \).
Since \( n > 0 \), \( n = 9 \).
Markscheme
(a)(i) Ratio: \( \frac{p \ln x}{\ln x} = \frac{\frac{1}{3} \ln x}{p \ln x} \implies p^2 = \frac{1}{3} \quad \mathbf{M1A1} \).
\( p = \pm \frac{1}{\sqrt{3}} \quad \mathbf{A1} \).
Note: Award M0A0 for \( p^2 = \frac{1}{3} \) with no working.
(a)(ii) \( |p| = \frac{1}{\sqrt{3}} < 1 \), so series converges \quad \mathbf{A1} \).
Note: Award R0 if both \( p \) values not considered.
(a)(iii) \( \frac{\ln x}{1 – \frac{1}{\sqrt{3}}} = 3 + \sqrt{3} \quad \mathbf{M1} \).
\( \ln x = 2 \implies x = e^2 \quad \mathbf{A1} \).
(b)(i) \( p – 1 = \frac{1}{3} – p \implies 2p = \frac{4}{3} \quad \mathbf{M1A1} \).
\( p = \frac{2}{3} \quad \mathbf{A1} \).
(b)(ii) \( d = -\frac{1}{3} \ln x \quad \mathbf{A1} \).
(b)(iii) \( S_n = \frac{n}{2} \left[ 2 \ln x + (n – 1) \left( -\frac{1}{3} \ln x \right) \right] = -3 \ln x \quad \mathbf{M1} \).
\( \frac{n}{2} \left( \frac{7 – n}{3} \right) = -3 \implies n^2 – 7n + 18 = 0 \quad \mathbf{M1A1} \).
\( n = 9 \quad \mathbf{A1} \).
[18 marks]
▶️ Answer
(a) First term \( a = 10 \), ratio \( r = \frac{5}{10} = 0.5 \).
General term: \( u_n = a \cdot r^{n-1} = 10 \cdot 0.5^{n-1} \).
(b) Find smallest \( u_n < 0.001 \): \( 10 \cdot 0.5^{n-1} < 0.001 \).
\( 0.5^{n-1} < 10^{-4} \implies (2^{-1})^{n-1} < 2^{-10} \implies -(n-1) < -10 \implies n > 11 \).
For \( n = 11 \): \( u_{11} = 10 \cdot 0.5^{10} = 0.0097656 > 0.001 \).
For \( n = 12 \): \( u_{12} = 10 \cdot 0.5^{11} \approx 0.0004883 < 0.001 \).
First term: \( 0.000488 \).
(c) Sum of first 20 terms: \( S_n = a \frac{1 – r^n}{1 – r} = 10 \cdot \frac{1 – 0.5^{20}}{1 – 0.5} \).
\( 0.5^{20} \approx 9.536 \times 10^{-7} \), so \( S_{20} = 10 \cdot \frac{1 – 9.536 \times 10^{-7}}{0.5} \approx 20 \cdot (1 – 9.536 \times 10^{-7}) \approx 19.999981 \).
To 6 decimal places: \( 19.999981 \).
(d) Infinite sum: \( S = \frac{a}{1 – r} = \frac{10}{1 – 0.5} = 20 \).
Markscheme
(a) \( u_n = 10 \cdot 0.5^{n-1} \) (or \( 20 \cdot 0.5^n \)) \quad \mathbf{A1} \).
(b) \( 10 \cdot 0.5^{n-1} < 0.001 \implies n > 11 \quad \mathbf{M1} \).
Test \( n = 12 \): \( u_{12} \approx 0.000488 < 0.001 \quad \mathbf{A1} \).
Answer: \( 0.000488 \quad \mathbf{A1} \).
(c) \( S_{20} = 10 \cdot \frac{1 – 0.5^{20}}{0.5} \approx 19.999981 \quad \mathbf{M1A1} \).
(d) \( S = \frac{10}{0.5} = 20 \quad \mathbf{A1} \).
[8 marks]
▶️ Answer
(i) First term \( a = 1 \), ratio \( r = \frac{\frac{2}{5}}{1} = \frac{2}{5} \).
Since \( |r| = \frac{2}{5} < 1 \), sum: \( S = \frac{a}{1 – r} = \frac{1}{1 – \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \).
(ii) First term \( a = 1 \), ratio \( r = \frac{-\frac{2}{5}}{1} = -\frac{2}{5} \).
Since \( |r| = \frac{2}{5} < 1 \), sum: \( S = \frac{a}{1 – r} = \frac{1}{1 – \left(-\frac{2}{5}\right)} = \frac{1}{1 + \frac{2}{5}} = \frac{1}{\frac{7}{5}} = \frac{5}{7} \).
Markscheme
(i) \( a = 1 \), \( r = \frac{2}{5} \), \( S = \frac{1}{1 – \frac{2}{5}} = \frac{5}{3} \quad \mathbf{M1A1A1} \).
(ii) \( a = 1 \), \( r = -\frac{2}{5} \), \( S = \frac{1}{1 – \left(-\frac{2}{5}\right)} = \frac{5}{7} \quad \mathbf{M1A1A1} \).
[6 marks]
Question:
[Maximum mark: 6]
Calculate the following sums using the appropriate formulas.
(i) \( \sum_{k=1}^{8} 4^k \)
(ii) \( \sum_{k=1}^{6} (0.25)^k \) (correct to 6 decimal places)
(iii) \( \sum_{k=1}^{\infty} (0.25)^k \)
▶️ Answer
(i) Finite geometric series: \( \sum_{k=1}^{8} 4^k \), first term \( a = 4 \), ratio \( r = 4 \), terms \( n = 8 \).
Sum: \( S_n = a \frac{r^n – 1}{r – 1} = 4 \cdot \frac{4^8 – 1}{4 – 1} = 4 \cdot \frac{65536 – 1}{3} = \frac{262140}{3} = 87380 \).
(ii) Finite geometric series: \( \sum_{k=1}^{6} (0.25)^k \), first term \( a = 0.25 \), ratio \( r = 0.25 \), terms \( n = 6 \).
Sum: \( S_n = a \frac{1 – r^n}{1 – r} = 0.25 \cdot \frac{1 – (0.25)^6}{1 – 0.25} = 0.25 \cdot \frac{1 – 0.000015625}{0.75} \approx 0.25 \cdot \frac{0.999984375}{0.75} \approx 0.333328125 \).
To 6 decimal places: \( 0.333328 \).
(iii) Infinite geometric series: \( \sum_{k=1}^{\infty} (0.25)^k \), first term \( a = 0.25 \), ratio \( r = 0.25 \).
Since \( |r| < 1 \), sum: \( S = \frac{a}{1 – r} = \frac{0.25}{1 – 0.25} = \frac{0.25}{0.75} = \frac{1}{3} \).
Markscheme
(i) \( S_8 = 4 \cdot \frac{4^8 – 1}{4 – 1} = 87380 \quad \mathbf{M1A1} \).
(ii) \( S_6 = 0.25 \cdot \frac{1 – (0.25)^6}{1 – 0.25} \approx 0.333328 \quad \mathbf{M1A1} \).
(iii) \( S = \frac{0.25}{1 – 0.25} = \frac{1}{3} \quad \mathbf{M1A1} \).
[6 marks]
▶️ Answer/Explanation
(a) First term \( a = 3 \), ratio \( r = 0.9 \). General term: \( u_n = a \cdot r^{n-1} \).
10th term: \( u_{10} = 3 \cdot (0.9)^{10-1} = 3(0.9)^9 \).
(b) Infinite sum: \( S = \frac{a}{1 – r} \).
Since \( |r| = 0.9 < 1 \), \( S = \frac{3}{1 – 0.9} = \frac{3}{0.1} = 30 \).
Markscheme
(a) \( u_{10} = 3(0.9)^9 \quad \mathbf{A1} \).
(b) Identify \( a = 3 \), \( r = 0.9 \quad \mathbf{M1} \).
Use \( S = \frac{a}{1 – r} \quad \mathbf{M1} \).
Compute: \( S = \frac{3}{1 – 0.9} = \frac{3}{0.1} \quad \mathbf{A1} \).
Answer: \( 30 \quad \mathbf{A1} \).
[5 marks]
▶️ Answer/Explanation
(a) Common ratio: \( r = \frac{5}{25} = \frac{1}{5} = 0.2 \).
(b)
(i) First term \( a = 25 \), ratio \( r = \frac{1}{5} \). 10th term: \( u_{10} = a \cdot r^{10-1} = 25 \cdot \left( \frac{1}{5} \right)^9 = 25 \cdot 5^{-9} = \frac{25}{5^9} = \frac{25}{1953125} = 0.0000128 \).
(ii) General term: \( u_n = a \cdot r^{n-1} = 25 \cdot \left( \frac{1}{5} \right)^{n-1} \).
(c) Infinite sum: \( S = \frac{a}{1 – r} \).
Since \( |r| = \frac{1}{5} < 1 \), \( S = \frac{25}{1 – \frac{1}{5}} = \frac{25}{\frac{4}{5}} = 25 \cdot \frac{5}{4} = \frac{125}{4} = 31.25 \).
Markscheme
(a) \( r = \frac{1}{5} \) (or 0.2) \quad \mathbf{A1} \).
(b)
(i) \( u_{10} = 25 \cdot \left( \frac{1}{5} \right)^9 = 0.0000128 \quad \mathbf{A1} \).
(ii) \( u_n = 25 \cdot \left( \frac{1}{5} \right)^{n-1} \quad \mathbf{A1} \).
(c) \( S = \frac{25}{1 – \frac{1}{5}} = \frac{125}{4} = 31.25 \quad \mathbf{M1A1} \).
[6 marks]