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3 years ago

Vince wrote the sequence below.

2 answers:

7 0

Let's rewrite the given sequence :1/3, 1, 5/3, 7/3 . Notice that 1 =3/3, then

1/3, 3/3, 5/3, 7/3,...We also notice that the common difference d is equal to:
3/3 - 1/3 = 2/3 and 5/3 - 3/3 = 2/3 & 7/3 - 5/3 = 2/3

Hence this sequence is NOT A GEOMETRIC PROGRESSION but an ARITHMETIC PROGRESSION,instead with d = 2/3

Oduvanchick [21]3 years ago

7 0

Answer:

The correct statement is:

<em>The sequence is not geometric because 2/3 was added to each term to get the next term.</em>

Step-by-step explanation:

Vince wrote the sequence below.

1/3, 1, 5/3, 7/3...

the sequence could also be written as:

1/3,3/3,5/3,7/3,.....

Let $a_n$ represents the nth term of the sequence.

now the first term of the sequence is:

$a_1=\dfrac{1}{3}$

second term is:

$a_2=\dfrac{3}{3}$

which could also be obtained as:

$a_2=a_1+\dfrac{2}{3}$

Third term is:

$a_3=\dfrac{5}{3}$

which could also be obtained as:

$a_3=a_2+\dfrac{2}{3}$

Fourth term is:

$a_4=\dfrac{7}{3}$

which could also be obtained as:

$a_4=a_3+\dfrac{2}{3}$

The correct reason that explains that the given sequence is geometric or not is:

The sequence is not geometric because 2/3 was added to each term to get the next term.

(Hence such a sequence is not geometric but an arithmetic sequence with common difference of 2/3)

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The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (

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Answer:

$\|a\| = 5\sqrt{13}$ .

$\|b\| = 3\sqrt{29}$ .

Step-by-step explanation:

Let $m$ , $n$ , and $k$ be scalars such that:

$\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}$ .

$\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}$ .

$\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}$ .

The question states that $\| a + b \| = 34$ . In other words:

$k\, \sqrt{8^{2} + 15^{2}} = 34$ .

$k^{2} \, (8^{2} + 15^{2}) = 34^{2}$ .

$289\, k^{2} = 34^{2}$ .

Make use of the fact that $289 = 17^{2}$ whereas $34 = 2 \times 17$ .

$\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}$ .

$k^{2} = 2^{2}$ .

The question also states that the scalar multiple here is positive. Hence, $k = 2$ .

Therefore:

$\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}$ .

$(a + b)$ could also be expressed in terms of $m$ and $n$ :

$\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}$ .

$\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}$ .

Equate the two expressions and solve for $m$ and $n$ :

$\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}$ .

$\begin{cases}m = 5 \\ n = 3\end{cases}$ .

Hence:

$\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}$ .

$\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}$ .

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