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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

(a) (i) To find the sum of the first five terms of the arithmetic sequence, we use the formula provided for \( S_n \):
\(
S_n = n^2 + 4n
\)
For \( n = 5 \), the sum of the first five terms, \( S_5 \), is calculated as:
\(
S_5 = 5^2 + 4(5)
\)
\(
S_5 = 25 + 20
\)
\(
S_5 = 45
\)
Therefore, the sum of the first five terms of the sequence is 45.

(ii)

Given that \( S_6 = 60 \), find \( u_6 \)**

We know:
\(
S_6 = 6^2 + 4(6)
\)
\(
S_6 = 36 + 24 = 60
\)
Now, to find \( u_6 \), we use the fact that:
\(
u_6 = S_6 – S_5
\)
\(
u_6 = 60 – 45
\)
\(
u_6 = 15
\)
So, \( u_6 = 15 \).

(b) The sum of the first \( n \) terms of an arithmetic sequence is given by:
\(
S_n = \frac{n}{2}\left(2u_1 + (n – 1)d\right)
\)
where \( u_1 \) is the first term and \( d \) is the common difference.
Given \( S_n = n^2 + 4n \), we can find \( u_1 \) by setting \( n = 1 \), which gives us the sum of just the first term \( u_1 \).
Therefore:
\(
S_1 = 1^2 + 4 \cdot 1
\)
\(
S_1 = 5
\)
Hence:
\(
u_1 = 5
\)

(c)  Firstly, we shall find the first term of the sequence, \( u_1 \). By setting \( n = 1 \) in the sum formula, we get:
\(
S_1 = 1^2 + 4(1) = 1 + 4 = 5
\)
Since \( S_1 \) represents the sum of the first term only, it follows that:
\(
u_1 = 5
\)
To find the common difference, \( d \), we can use the formula for the sum of the first \( n \) terms again with \( n = 2 \):
\(
S_2 = 2^2 + 4(2) = 4 + 8 = 12
\)
The sum of the first two terms is the sum of \( u_1 \) and \( u_2 \), hence:
\(
u_2 = S_2 – u_1 = 12 – 5 = 7
\)
The common difference \( d \) is then:
\(
u_2 – u_1 = 7 – 5 = 2
\)
Having established that \( u_1 = 5 \) and \( d = 2 \), we can now express the \( n^\text{th} \) term of the sequence using the formula:
\(
u_n = u_1 + (n – 1)d
\)
\(
u_n = 5 + (n – 1)(2)
\)
Simplifying this, we get:
\(
u_n = 5 + 2n – 2
\)
\(
u_n = 2n + 3
\)

(d) The first term \( u_1 \) is simply the sum of the first term, so:
\(
S_1 = u_1 = 1^2 + 4 \times 1 = 5
\)
Therefore:
\(
v_2 = 5
\)
To find \( u_6 \), we calculate \( S_6 – S_5 \) since \( S_n \) is the sum of the first \( n \) terms.
The sixth term can be found by subtracting the sum of the first five terms from the sum of the first six terms:
\(
S_6 = 6^2 + 4 \times 6 = 36 + 24 = 60
\)
and:
\(
S_5 = 5^2 + 4 \times 5 = 25 + 20 = 45
\)
Thus:
\(
u_6 = S_6 – S_5 = 60 – 45 = 15
\)
and hence:
\(
v_4 = 15
\)
Now, we have \( v_2 \) and \( v_4 \) of the geometric sequence; we can use these to find the common ratio \( r \).
In a geometric sequence, the \( n \)-th term \( v_n \) is given by:
\(
v_n = v_1 r^{n-1}
\)
Therefore:
\(
v_4 = v_2 r^2
\)
Substituting the known values gives us:
\(
15 = 5r^2
\)
which simplifies to:
\(
r^2 = 3
\)
Finally, solving for \( r \) gives us:
\(
r = \pm \sqrt{3}
\)

(e) The sum of the first \( n \) terms of \( (u_n) \) is given by:
\(
S_n = n^2 + 4n
\)
from which we can find:
\(
u_1 = S_1 = 5 \quad\text{and}\quad u_6 = S_6 – S_5 = 15
\)
Knowing that \( v_2 = 5 \) and \( v_4 = 15 \), and using the property of geometric sequences \( v_n = ar^{(n-1)} \), we establish the system:
\(
v_2 = ar = 5 \quad\text{and}\quad v_4 = ar^3 = 15
\)
Dividing the latter by the former yields:
\(
r^2 = 3
\)
hence:
\(
r = \pm\sqrt{3}
\)
Given \( v_{99} < 0 \), we deduce:
\(
r = -\sqrt{3}
\)
To find \( v_5 \), we calculate:
\(
v_5 = \left(\frac{5}{-\sqrt{3}}\right) \times \left(-\sqrt{3}\right)^4
\)
\(
= -15\sqrt{3}
\)

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