Home / IB MYP Year 4-5: Exntended Mathematics : Unit 2: Algebra -Arithmetic and geometric sequences MYP Style Questions

IB MYP Year 4-5: Exntended Mathematics : Unit 2: Algebra -Arithmetic and geometric sequences MYP Style Questions

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

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