IB myp 4-5 MATHEMATICS – Practice Questions- All Topics
Topic :Algebra-Airthemetic and Geometric Sequence
Topic :Algebra- Weightage : 21 %
All Questions for Topic : Airthemetic and Geometric Sequence
Question
Below is a geometric sequence with first term $U_1=4$ and common ratio $r$
$4,2 \sqrt{2}, 2, \ldots$
Question (a)
Write down the value of $r$.
▶️Answer/Explanation
Ans:
In a geometric sequence, each term is obtained by multiplying the previous term by a constant factor called the common ratio, denoted by \(r\). In this case, we are given the first term \(U_1 = 4\) and the second term \(U_2 = 2 \sqrt{2}\).
The relationship between consecutive terms in a geometric sequence is given by:
\[U_{n+1} = U_n \cdot r\]
Using this relationship, we can find the common ratio \(r\) by dividing the second term by the first term:
\[r = \frac{U_2}{U_1} = \frac{2 \sqrt{2}}{4} = \frac{\sqrt{2}}{2}\]
So, the value of the common ratio \(r\) is \(\frac{\sqrt{2}}{2}\).
Question (b)
By continuing the pattern, determine the value of $n$ when $U_n=r$.
▶️Answer/Explanation
Ans:
We know that in a geometric sequence, the \(n\)th term can be expressed as:
\[U_n = U_1 \cdot r^{n-1}\]
Given that \(U_1 = 4\) and \(r = \frac{\sqrt{2}}{2}\), we want to find the value of \(n\) when \(U_n = r\). Substituting the values:
\[r = U_1 \cdot r^{n-1}\]
\[\frac{\sqrt{2}}{2} = 4 \cdot \left(\frac{\sqrt{2}}{2}\right)^{n-1}\]
Dividing both sides by \(4\):
\[\left(\frac{\sqrt{2}}{2}\right)^{n-1} = \frac{1}{4\sqrt 2}\]
Taking the natural logarithm of both sides:
\[\ln\left(\left(\frac{1}{\sqrt 2}\right)^{n-1}\right) = \ln\left(\frac{1}{\sqrt {2} ^4 \times \sqrt 2}\right)\]
Using the logarithm property \(\log_a(b^c) = c \cdot \log_a(b)\):
\[n-1=5\Rightarrow n=6\]
Question (c)
Given that $U_{21}=2^k ; k \in Z$, find the value of $k$.
▶️Answer/Explanation
Ans:
In a geometric sequence, the \(n\)th term \(U_n\) can be expressed in terms of the first term \(U_1\) and the common ratio \(r\) as follows:
\[U_n = U_1 \cdot r^{n-1}\]
In this case, you’ve mentioned that \(U_{21} = 2^k\), and we need to find the value of \(k\). The \(n\)th term is \(U_{21}\), the first term \(U_1 = 4\), and the common ratio \(r = \frac{\sqrt{2}}{2}\).
Plugging these values into the formula:
\[U_{21} = 4 \cdot \left(\frac{\sqrt{2}}{2}\right)^{21-1}\]
Simplifying the exponent:
\[U_{21} = 4 \cdot \left(\frac{\sqrt{2}}{2}\right)^{20} \Rightarrow 4 \cdot \frac{2^{10}}{2^{20}} \Rightarrow\frac{ 2^{12}}{2^{20}}\]
Since \(U_{21} = 2^k\), we have \(2^{-8} = 2^k\), which implies that \(k = -8\).
Therefore, the value of \(k\) is -8