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SL 1.3 Geometric sequences and series HL Paper 1 – IBDP Maths AA HL- Exam Style Questions

IBDP Maths AA HL Topic - SL 1.3 Geometric sequences and series HL Paper 1- Exam Style Questions

IB MATHEMATICS AA HL – Practice Questions- Paper1-All Topics

Topic : SL 1.3 Geometric sequences and series HL Paper 1

Topic 1- Number and algebra- Weightage : 15 % 

All Questions for Topic : SL 1.3 –Geometric sequences and series Use of the formulae for the nth term and the sum of the first n terms of the sequence. Use of sigma notation for the sums of geometric sequences. Applications .Examples include the spread of disease, salary increase and decrease and population growth

Question
Consider a geometric sequence \( \{u_n\} \) satisfying \( u_3 = 9 \) and \( S_3 = u_1 + u_2 + u_3 = 13 \).

Find the first term \( u_1 \) and the common ratio \( r \).
▶️ Answer/Explanation
Solution
Let the first term be \( u_1 = a \), and the common ratio be \( r \).
Then the terms are:
\( u_2 = ar \),
\( u_3 = ar^2 \).
Given \( u_3 = 9 \Rightarrow ar^2 = 9 \quad \text{(1)} \)
Also \( S_3 = a + ar + ar^2 = 13 \quad \text{(2)} \)
From (1): \( a = \frac{9}{r^2} \)
Substitute into (2):
\[ \frac{9}{r^2} + \frac{9r}{r^2} + 9 = 13 \]
Simplify:
\[ \frac{9(1 + r)}{r^2} + 9 = 13 \]
\[ \frac{9(1 + r)}{r^2} = 4 \Rightarrow 9(1 + r) = 4r^2 \Rightarrow 4r^2 – 9r – 9 = 0 \]
Solve the quadratic:
\[ r = \frac{9 \pm \sqrt{81 + 144}}{8} = \frac{9 \pm \sqrt{225}}{8} = \frac{9 \pm 15}{8} \]
\[ \Rightarrow r = 3 \quad \text{or} \quad r = -\frac{3}{4} \]
If \( r = 3 \), then \( a = \frac{9}{9} = 1 \)
If \( r = -\frac{3}{4} \), then \( a = \frac{9}{\left(-\frac{3}{4}\right)^2} = \frac{9}{\frac{9}{16}} = 16 \)
✅ Answers:
If \( r = 3 \), then \( u_1 = 1 \)
If \( r = -\frac{3}{4} \), then \( u_1 = 16 \)
Question

Consider the arithmetic sequence \(u_{1} , u_{2} , u_{3}\) , ….
The sum of the first n terms of this sequence is given by \(S_{n} = n^2+4n\)
(a) (i) Find the sum of the first five terms.
(ii) Given that \(S_{6}\) = 60, find \(u_{6}\) .
(b) Find \(u_{1}\) .
(c) Hence or otherwise, write an expression for \(u_{n}\) in terms of n.
Consider a geometric sequence, \(v_{n}\) , where \(v_{2}\) = \(u_{1}\) and \(v_{4}\) = \(u_{6}\) .
(d) Find the possible values of the common ratio, r.
(e) Given that \(v_{99}\) < 0, find \(v_{5}\)

▶️ Answer/Explanation
Solution

(a)(i)
\( S_n = n^2 + 4n \).
For \( n = 5 \): \( S_5 = 5^2 + 4(5) = 25 + 20 = 45 \).

(a)(ii)
Given \( S_6 = 60 \), \( S_5 = 45 \).
\( u_6 = S_6 – S_5 = 60 – 45 = 15 \).

(b)
For \( n = 1 \): \( S_1 = 1^2 + 4(1) = 5 \).
Thus, \( u_1 = 5 \).

(c)
\( u_n = S_n – S_{n-1} \).
\( S_{n-1} = (n-1)^2 + 4(n-1) = n^2 + 2n – 3 \).
\( u_n = (n^2 + 4n) – (n^2 + 2n – 3) = 2n + 3 \).

(d)
Given: \( v_2 = u_1 = 5 \), \( v_4 = u_6 = 15 \).
For geometric sequence: \( v_2 = v_1 r \), \( v_4 = v_1 r^3 \).
\( \frac{v_4}{v_2} = r^2 = \frac{15}{5} = 3 \).
Thus, \( r = \pm \sqrt{3} \).

(e)
Given \( v_{99} < 0 \), \( v_{99} = v_1 r^{98} \).
Since \( r^{98} > 0 \), \( v_1 < 0 \).
From \( v_2 = v_1 r = 5 \), if \( r = -\sqrt{3} \), \( v_1 = \frac{5}{-\sqrt{3}} \).
\( v_5 = v_1 (-\sqrt{3})^4 = \frac{5}{-\sqrt{3}} \cdot 9 = -15\sqrt{3} \).

Markscheme
(a)(i) \( S_5 = 5^2 + 4(5) = 45 \).
(a)(ii) \( S_6 = 60 \), \( S_5 = 45 \), \( u_6 = 60 – 45 = 15 \).
(b) \( S_1 = 1^2 + 4(1) = 5 \), so \( u_1 = 5 \).
(c) \( u_2 = S_2 – S_1 = 12 – 5 = 7 \), \( d = 7 – 5 = 2 \), so \( u_n = 5 + (n-1)2 = 2n + 3 \).
(d) \( v_2 = 5 \), \( v_4 = 15 \), \( v_4 = v_2 r^2 \), \( 15 = 5r^2 \), \( r^2 = 3 \), \( r = \pm \sqrt{3} \).
(e) \( v_{99} < 0 \implies r = -\sqrt{3} \), \( v_1 = \frac{5}{-\sqrt{3}} \), \( v_5 = \frac{5}{-\sqrt{3}} (-\sqrt{3})^4 = -15\sqrt{3} \).

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