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Inga [223]
2 years ago
9

Find the value of the term in the arithmetic sequence using an = a^1+ (n-1)d. 3,5,7,9,11, (8th term)

Mathematics
2 answers:
saveliy_v [14]2 years ago
8 0
<h3>Answer:  17</h3>

=========================================================

Work Shown:

a1 = 3 = first term

d = 2 = common difference (since we add 2 to each term to get the next one)

Let's compute the nth term.

an = a1 + (n-1)*d

an = 3 + (n-1)*2

an = 3 + 2n-2

an = 2n+1

-----------------------

To check things so far, we can plug in something like n = 2

an = 2n+1

a2 = 2*2+1

a2 = 5

Showing that the 2nd term is 5, which matches with the sequence given to us

Let's check n = 3

an = 2n+1

a3 = 2*3+1

a3 = 7

That matches as well. I'll let you check the others.

-----------------------

Plug in n = 8 to find the 8th term

an = 2n+1

a8 = 2*8+1

a8 = 17

The eighth term is 17, which is the final answer.

-----------------------

You could extend out the given sequence by adding 2 each time until you reach the 8th term

3,5,7,9,11,13,15,17

Though this method is slow if you need to find say the 38th term

Vesna [10]2 years ago
6 0

Answer:

17 is the answer for the value of the term in the arithmetic sequence using the above

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\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})

We have been given two vectors $\vec{a}$ and $\vec{b}$, we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of$\vec{b}$onto $\vec{a}$means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

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