IB Mathematics SL 1.3 Geometric sequences and series AA SL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a)(i)
Recognize geometric sequence: ratio of consecutive terms is constant.
Correct expression for common ratio:
\( r = \frac{6}{m – 1} \) or \( r = \frac{m + 4}{6} \) (A1 N1)
[2 marks]
Part (a)(ii)
Equate common ratio expressions:
\( \frac{6}{m – 1} = \frac{m + 4}{6} \) (A1)
Cross-multiply:
\( 6 \cdot 6 = (m – 1)(m + 4) \)
\( 36 = m^2 + 3m – 4 \) (A1)
Rearrange to required form:
\( m^2 + 3m – 4 – 36 = 0 \)
\( m^2 + 3m – 40 = 0 \) (A1 AG N0)
[2 marks]
Part (b)(i)
Valid attempt to solve quadratic equation \( m^2 + 3m – 40 = 0 \):
\( (m + 8)(m – 5) = 0 \) or use quadratic formula \( m = \frac{-3 \pm \sqrt{9 + 160}}{2} \) (M1)
Solutions:
\( m = -8, 5 \) (A1 A1 N3)
[3 marks]
Part (b)(ii)
METHOD 1
Substitute \( m \) into \( r = \frac{6}{m – 1} \):
For \( m = -8 \): \( r = \frac{6}{-8 – 1} = -\frac{6}{9} = -\frac{2}{3} \)
For \( m = 5 \): \( r = \frac{6}{5 – 1} = \frac{6}{4} = \frac{3}{2} \) (M1 A1 A1)
METHOD 2
Substitute \( m \) into \( r = \frac{m + 4}{6} \):
For \( m = -8 \): \( r = \frac{-8 + 4}{6} = -\frac{4}{6} = -\frac{2}{3} \)
For \( m = 5 \): \( r = \frac{5 + 4}{6} = \frac{9}{6} = \frac{3}{2} \) (M1 A1 A1)
Possible values: \( r = \frac{3}{2}, -\frac{2}{3} \) (N3)
[3 marks]
Part (c)(i)
State: \( r = -\frac{2}{3} \) (A1)
Justify: For an infinite geometric sequence to have a finite sum, the common ratio must satisfy \( |r| < 1 \).
\( \left| -\frac{2}{3} \right| = \frac{2}{3} < 1 \), whereas \( \left| \frac{3}{2} \right| = \frac{3}{2} > 1 \). Thus, only \( r = -\frac{2}{3} \) yields a finite sum (R1 N0)
[2 marks]
Part (c)(ii)
Find first term for \( m = -8 \) (since \( r = -\frac{2}{3} \)):
\( u_1 = m – 1 = -8 – 1 = -9 \) (A1)
Substitute into infinite sum formula \( S_\infty = \frac{u_1}{1 – r} \):
\( S_\infty = \frac{-9}{1 – \left(-\frac{2}{3}\right)} = \frac{-9}{1 + \frac{2}{3}} = \frac{-9}{\frac{5}{3}} \) (A1)
Calculate:
\( S_\infty = -9 \cdot \frac{3}{5} = -\frac{27}{5} = -5.4 \) (A1 N3)
[3 marks]
Total [15 marks]
▶️ Answer/Explanation
Part (a)
Solve cubic equation \( 4p^3 + 4p^2 – 15p – 8 = 0 \):
Use rational root theorem; test possible roots \( \pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}, \pm \frac{1}{4} \).
Test \( p = -\frac{1}{2} \): \( 4\left(-\frac{1}{8}\right) + 4\left(\frac{1}{4}\right) – 15\left(-\frac{1}{2}\right) – 8 = -\frac{1}{2} + 1 + \frac{15}{2} – 8 = 0 \) (M1)
Factor using synthetic division with \( p = -\frac{1}{2} \):
Quotient: \( 4p^2 + 2p – 16 \), so \( 4p^3 + 4p^2 – 15p – 8 = (p + \frac{1}{2})(4p^2 + 2p – 16) \).
Simplify: \( (2p + 1)(2p^2 + p – 8) = 0 \).
Solve quadratic: \( 2p^2 + p – 8 = 0 \), discriminant \( \Delta = 1 + 64 = 65 \),
\( p = \frac{-1 \pm \sqrt{65}}{4} \approx 1.765, -2.265 \) (A1)
Roots: \( p = -\frac{1}{2}, \frac{-1 + \sqrt{65}}{4}, \frac{-1 – \sqrt{65}}{4} \) (A1 N3)
[3 marks]
Part (b)
Geometric sequence: first term \( u_1 = 4 \), common ratio \( p \).
General term: \( u_n = 4 \cdot p^{n-1} \).
First four terms: \( u_1 = 4 \), \( u_2 = 4p \), \( u_3 = 4p^2 \), \( u_4 = 4p^3 \) (A1 A1 N2)
[2 marks]
Part (c)
Arithmetic sequence: first term \( a_1 = 4 \), common difference \( d = 5p \).
General term: \( a_n = 4 + (n-1) \cdot 5p \).
First four terms: \( a_1 = 4 \), \( a_2 = 4 + 5p \), \( a_3 = 4 + 10p \), \( a_4 = 4 + 15p \) (A1 A1 N2)
[2 marks]
Part (d)
Sum of third and fourth terms of geometric sequence: \( u_3 + u_4 = 4p^2 + 4p^3 \).
Sum of first and fourth terms of arithmetic sequence: \( a_1 + a_4 = 4 + (4 + 15p) = 8 + 15p \).
Equate: \( 4p^2 + 4p^3 = 8 + 15p \) (M1)
Rearrange: \( 4p^3 + 4p^2 – 15p – 8 = 0 \).
This is the same cubic as in Part (a), so solutions are:
\( p = -\frac{1}{2}, \frac{-1 + \sqrt{65}}{4}, \frac{-1 – \sqrt{65}}{4} \) (A1 A1 N3)
[3 marks]
Part (e)
Geometric series sum to infinity exists if \( |p| < 1 \).
Check roots: \( \left| -\frac{1}{2} \right| = \frac{1}{2} < 1 \), \( \left| \frac{-1 + \sqrt{65}}{4} \right| \approx 1.765 > 1 \), \( \left| \frac{-1 – \sqrt{65}}{4} \right| \approx 2.265 > 1 \).
Thus, \( p = -\frac{1}{2} \) (A1)
Use infinite sum formula: \( S_\infty = \frac{u_1}{1 – p} \).
Substitute \( u_1 = 4 \), \( p = -\frac{1}{2} \):
\( S_\infty = \frac{4}{1 – \left(-\frac{1}{2}\right)} = \frac{4}{1 + \frac{1}{2}} = \frac{4}{\frac{3}{2}} = 4 \cdot \frac{2}{3} = \frac{8}{3} \) (M1 A1)
Sum to infinity: \( \frac{8}{3} \approx 2.66667 \) (A1 N2)
[4 marks]
Total [14 marks]
▶️ Answer/Explanation
Part (a)
Apply logarithm property: \( \ln{x^n} = n \ln x \).
Terms: \( \ln{x^{16}} = 16 \ln x \), \( \ln{x^8} = 8 \ln x \), \( \ln{x^4} = 4 \ln x \) (A1)
Find common ratio \( r \):
\( r = \frac{u_2}{u_1} = \frac{8 \ln x}{16 \ln x} = \frac{8}{16} = \frac{1}{2} \)
Or: \( r = \frac{u_3}{u_2} = \frac{4 \ln x}{8 \ln x} = \frac{4}{8} = \frac{1}{2} \) (M1)
Common ratio: \( r = \frac{1}{2} \) (A1 N2)
[3 marks]
Part (b)
Method 1
Recognize the sum as an infinite geometric series:
Sum: \( \sum_{k=1}^\infty 2^{5-k} \ln x \).
Rewrite terms: \( 2^{5-k} = 2^5 \cdot 2^{-k} = 32 \cdot \left(\frac{1}{2}\right)^k \).
Sum: \( \sum_{k=1}^\infty 32 \cdot \left(\frac{1}{2}\right)^k \cdot \ln x \) (M1)
Identify geometric series: first term \( a = 32 \ln x \cdot \left(\frac{1}{2}\right)^1 = 16 \ln x \), common ratio \( r = \frac{1}{2} \) (from Part (a)).
Since \( \left| \frac{1}{2} \right| < 1 \), use infinite sum formula: \( S_\infty = \frac{a}{1 – r} \) (M1)
Substitute: \( S_\infty = \frac{16 \ln x}{1 – \frac{1}{2}} = \frac{16 \ln x}{\frac{1}{2}} = 32 \ln x \).
Set equal to 64: \( 32 \ln x = 64 \) (A1)
Solve: \( \ln x = \frac{64}{32} = 2 \)
\( x = e^2 \approx 7.389 \) (A1 A1 N3)
Method 2
Rewrite sum using logarithm properties:
\( 2^{5-k} \ln x = \ln \left( x^{2^{5-k}} \right) \).
Sum: \( \sum_{k=1}^\infty \ln \left( x^{2^{5-k}} \right) = \ln \left( \prod_{k=1}^\infty x^{2^{5-k}} \right) \).
Exponent: \( \sum_{k=1}^\infty 2^{5-k} = 2^4 + 2^3 + 2^2 + \cdots = 16 \cdot \frac{1}{1 – \frac{1}{2}} = 32 \) (M1)
So: \( \ln \left( x^{32} \right) = 32 \ln x \).
Set equal to 64: \( 32 \ln x = 64 \) (A1)
Solve: \( \ln x = 2 \)
\( x = e^2 \) (A1 A1 N3)
[5 marks]
Total [8 marks]