IB Mathematics SL 1.2 Arithmetic Sequences & Series AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a) [2 marks]
Arithmetic sequence formula: \( u_n = u_1 + (n-1)d \).
For \( u_4 \): \( u_4 = u_1 + 3d \).
Given: \( u_1 = 0.6 \), \( u_4 = 0.15 \).
Substitute: \( 0.15 = 0.6 + 3d \).
Solve: \( 3d = 0.15 – 0.6 = -0.45 \).
\[ d = \frac{-0.45}{3} = -0.15 \]
Thus, the common difference is \( d = -0.15 \).
Part (b) [4 marks]
Probability sum: \( \sum P(X = n) = 1 \).
Given: \( P(X = n) = \frac{u_n}{k} \), \( n = 1, 2, 3, 4 \).
Sequence: \( u_1 = 0.6 \), \( d = -0.15 \).
\[ u_2 = 0.6 – 0.15 = 0.45 \]
\[ u_3 = 0.45 – 0.15 = 0.3 \]
\[ u_4 = 0.3 – 0.15 = 0.15 \]
Sum: \( \frac{u_1}{k} + \frac{u_2}{k} + \frac{u_3}{k} + \frac{u_4}{k} = 1 \).
\[ \frac{0.6 + 0.45 + 0.3 + 0.15}{k} = 1 \]
\[ \frac{1.5}{k} = 1 \]
\[ k = 1.5 \]
Alternatively, using the sum of an arithmetic sequence: \( S_4 = \frac{4}{2} (u_1 + u_4) = \frac{4}{2} (0.6 + 0.15) = 2 \times 0.75 = 1.5 \).
Thus, \( k = 1.5 \).
(a)
\( u_1 + 3d = u_4 \) (M1)
\( 0.6 + 3d = 0.15 \)
\( d = -0.15 \) (A1)
(b)
METHOD 1
\( u_2 = 0.45 \) or \( u_3 = 0.3 \) (may be seen in their equation) (A1)
Summing their probabilities to 1 (seen anywhere) (M1)
\( \frac{0.6}{k} + \frac{u_2}{k} + \frac{u_3}{k} + \frac{0.15}{k} = 1 \)
\( 0.6 + 0.45 + 0.3 + 0.15 = 1 \) (or equivalent) (A1)
\( \frac{1.5}{k} = 1 \)
\( k = 1.5 \) (A1)
METHOD 2 (using \( S_n \) formula)
\( S_4 = \frac{4}{2}(2(0.6)+(4-1)(-0.15)) \) OR \( S_4 = \frac{4}{2}(0.6+0.15) \) (or equivalent) (A1)
Summing their probabilities to 1 (seen anywhere) (M1)
\( \frac{u_1}{k} + \frac{u_2}{k} + \frac{u_3}{k} + \frac{u_4}{k} = 1 \) OR \( u_1 + u_2 + u_3 + u_4 = k \)
\( \frac{4}{2}(2(0.6)+(4-1)(-0.15)) = k \) (or equivalent) (A1)
\( k = 1.5 \) (A1)
▶️ Answer/Explanation
Part (a) [2 marks]
In an arithmetic sequence, the \( n \)-th term is given by \( u_n = a + (n-1)d \), where \( a \) is the first term and \( d \) is the common difference.
Given: second term \( u_2 = a + d = 10 \) (1), fourth term \( u_4 = a + 3d = 22 \) (2).
Subtract equation (1) from equation (2):
\[ (a + 3d) – (a + d) = 22 – 10 \]
\[ 2d = 12 \]
\[ d = 6 \]
Alternatively:
\[ d = \frac{u_4 – u_2}{4 – 2} = \frac{22 – 10}{2} = \frac{12}{2} = 6 \]
Thus, the common difference is \( d = 6 \).
Part (b) [2 marks]
From part (a), \( d = 6 \). Use the second term to find the first term \( a \):
\[ u_2 = a + d = 10 \]
\[ a + 6 = 10 \]
\[ a = 4 \]
The \( n \)-th term formula is:
\[ u_n = a + (n-1)d \]
Substitute \( a = 4 \), \( d = 6 \):
\[ u_n = 4 + (n-1) \cdot 6 \]
\[ u_n = 4 + 6n – 6 = 6n – 2 \]
Thus, \( u_n = 6n – 2 \).
(a)
Valid method to find the common difference: \( d = \frac{22-10}{2} \) or \( 10 + 2d = 22 \) or \( u_1 + d = 10 \), \( u_1 + 3d = 22 \) or \( u_3 = 16 \) (M1)
\[ d = 6 \] (A1)
(b)
First term: \( u_1 = 10 – 6 = 4 \) (M1)
Expression: \( u_n = 4 + 6(n-1) \) or \( u_n = 6n – 2 \) (A1)
▶️ Answer/Explanation
Part (a) [3 marks]
(i) The sum of the first \( n \) terms is given by \( S_n = n^2 + 4n \). For \( n = 5 \):
\[ S_5 = 5^2 + 4 \cdot 5 = 25 + 20 = 45 \]
Thus, the sum of the first five terms is 45.
(ii) Given \( S_6 = 60 \), find \( u_6 \). Use the relationship \( S_6 = S_5 + u_6 \):
Calculate \( S_6 \): \( S_6 = 6^2 + 4 \cdot 6 = 36 + 24 = 60 \), which is given.
Calculate \( S_5 \) from part (i): \( S_5 = 45 \).
\[ u_6 = S_6 – S_5 = 60 – 45 = 15 \]
Alternatively, use the sum formula: \( S_n = \frac{n}{2} (u_1 + u_n) \). For \( n = 6 \):
\[ S_6 = \frac{6}{2} (u_1 + u_6) = 3 (u_1 + u_6) = 60 \]
\[ u_1 + u_6 = 20 \]
From part (b) below, \( u_1 = 5 \), so:
\[ 5 + u_6 = 20 \]
\[ u_6 = 15 \]
Thus, \( u_6 = 15 \).
Part (b) [1 mark]
The first term is \( u_1 = S_1 \):
\[ S_1 = 1^2 + 4 \cdot 1 = 1 + 4 = 5 \]
Thus, \( u_1 = 5 \).
Part (c) [2 marks]
The \( n \)-th term is found by \( u_n = S_n – S_{n-1} \):
\[ S_n = n^2 + 4n \]
\[ S_{n-1} = (n-1)^2 + 4(n-1) = (n^2 – 2n + 1) + 4n – 4 = n^2 + 2n – 3 \]
\[ u_n = S_n – S_{n-1} = (n^2 + 4n) – (n^2 + 2n – 3) = 2n + 3 \]
Alternatively, use the arithmetic sequence formula \( u_n = u_1 + (n-1)d \). Find \( d \):
\[ u_2 = S_2 – S_1 \], where \( S_2 = 2^2 + 4 \cdot 2 = 4 + 8 = 12 \), \( S_1 = 5 \).
\[ u_2 = 12 – 5 = 7 \]
\[ d = u_2 – u_1 = 7 – 5 = 2 \]
With \( u_1 = 5 \), \( d = 2 \):
\[ u_n = 5 + (n-1) \cdot 2 = 5 + 2n – 2 = 2n + 3 \]
Thus, \( u_n = 2n + 3 \).
Part (d) [2 marks]
The geometric sequence \( v_n \) has \( v_2 = u_1 = 5 \) and \( v_4 = u_6 = 15 \). For a geometric sequence, \( v_n = v_1 r^{n-1} \), so:
\[ v_2 = v_1 r = 5 \]
\[ v_4 = v_1 r^3 = 15 \]
Divide the equations:
\[ \frac{v_4}{v_2} = \frac{v_1 r^3}{v_1 r} = r^2 \]
\[ r^2 = \frac{15}{5} = 3 \]
\[ r = \pm \sqrt{3} \]
Thus, the possible values of the common ratio are \( r = \sqrt{3} \) or \( r = -\sqrt{3} \).
Part (e) [2 marks]
Given \( v_{99} < 0 \), determine the sign of \( r \). For a geometric sequence, \( v_n = v_1 r^{n-1} \). Since \( v_{99} = v_1 r^{98} \), and \( r^{98} = (r^2)^{49} \), if \( r^2 = 3 \), then \( r^{98} > 0 \). Thus, the sign of \( v_{99} \) depends on \( v_1 \).
From \( v_2 = v_1 r = 5 \), if \( r = -\sqrt{3} \):
\[ v_1 = \frac{5}{-\sqrt{3}} = -\frac{5}{\sqrt{3}} = -\frac{5\sqrt{3}}{3} \]
\[ v_{99} = v_1 (-\sqrt{3})^{98} = -\frac{5\sqrt{3}}{3} \cdot (\sqrt{3})^{98} = -\frac{5\sqrt{3}}{3} \cdot 3^{49} < 0 \]
If \( r = \sqrt{3} \), then \( v_1 = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \), and \( v_{99} = \frac{5\sqrt{3}}{3} \cdot 3^{49} > 0 \), which does not satisfy \( v_{99} < 0 \). Thus, \( r = -\sqrt{3} \).
Calculate \( v_5 \):
\[ v_5 = v_1 (-\sqrt{3})^4 = -\frac{5\sqrt{3}}{3} \cdot (\sqrt{3})^4 = -\frac{5\sqrt{3}}{3} \cdot 9 = -15\sqrt{3} \]
Thus, \( v_5 = -15\sqrt{3} \).
(a)
(i) Recognition that \( n = 5 \): \( S_5 = 5^2 + 4 \cdot 5 \) (M1)
\[ S_5 = 45 \] (A1)
(ii) Method 1: Recognition that \( S_5 + u_6 = S_6 \) (M1)
\[ u_6 = 15 \] (A1)
Or Method 2: Recognition that \( 60 = \frac{6}{2} (u_1 + u_6) \) (M1)
\[ u_6 = 15 \] (A1)
Or Method 3: Substituting \( u_1 \) and \( d \) into \( u_1 + (n-1)d \) (M1)
\[ u_6 = 15 \] (A1)
(b)
Recognition that \( u_1 = S_1 \) or substituting \( u_6 \) into \( S_6 \): \( S_1 = 1 + 4 \) or \( 60 = \frac{6}{2} (u_1 + 15) \) (M1)
\[ u_1 = 5 \] (A1)
(c)
Either: Valid attempt to find \( d \): \( d = u_2 – u_1 \), \( d = 2 \) (M1)
Or: Valid attempt to find \( S_n – S_{n-1} \): \( (n^2 + 4n) – (n^2 – 2n + 1 + 4n – 4) \) (M1)
Or: Equating \( n^2 + 4n = \frac{n}{2} (5 + u_n) \): \( 2n + 8 = 5 + u_n \) (M1)
\[ u_n = 2n + 3 \] or \( u_n = 5 + 2(n-1) \) (A1)
(d)
Recognition that \( v_2 r^2 = v_4 \): \( r^2 = \frac{v_4}{v_2} = \frac{15}{5} = 3 \) (M1)
\[ r = \pm \sqrt{3} \] (A1)
(e)
Recognition that \( r \) is negative: \( r = -\sqrt{3} \) (M1)
\[ v_5 = -15\sqrt{3} \] (A1)