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
Find the first term \( u_1 \) and the common ratio \( r \).
▶️ Answer/Explanation
\( u_2 = ar \),
\( u_3 = ar^2 \).
\[ \frac{9}{r^2} + \frac{9r}{r^2} + 9 = 13 \]
\[ \frac{9(1 + r)}{r^2} + 9 = 13 \]
\[ r = \frac{9 \pm \sqrt{81 + 144}}{8} = \frac{9 \pm \sqrt{225}}{8} = \frac{9 \pm 15}{8} \]
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:
\(
v_5 = \left(\frac{5}{-\sqrt{3}}\right) \times \left(-\sqrt{3}\right)^4
\)
\(
= -15\sqrt{3}
\)