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3 years ago

Write an equation for the nth term of the arithmetic sequence: 15,28,41 ....

Mathematics

1 answer:

11111nata11111 [884]3 years ago

4 0

Answer:

The equation for the nth term of the given arithmetic sequence is: $\mathbf{a_n=13n+22}$

Step-by-step explanation:

We need to write an equation for the nth term of the arithmetic sequence:

15,28,41 ....

The equation for arithmetic sequence is: $a_n=a_1+(n-1)d$

Where $a_n$ is the nth term, $a_1$ is first term and d is common difference

In the given sequence we have:

a₁ = 15

a₂ = 28

We can find common difference using the formula:

$a_n=a_1+(n-1)d\\Put\: n=2, a_2=28\: and\: a_1=15\\a_2=a_1+(2-1)d\\28=15+d\\d=28-15\\d=13$

So, the common difference d is 13

Now, equation for nth term will be:

$a_n=a_1+(n-1)d\\Put\:a_1=15, d=13\\a_n=15+(n-1)13\\Solving:\\a_n=15+13n-13\\a_n=13n+2$

So, the equation for the nth term of the given arithmetic sequence is: $\mathbf{a_n=13n+22}$

where n=1,2,3..

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Question 3(Multiple Choice Worth 2 points)

Lilit [14]

The equivalent expression is seven-fifths squared times one-third cubed. Then the correct option is D.

<h3>What is an equivalent expression?</h3>

The equivalent is the expression that is in different forms but is equal to the same value.

Five-sevenths squared times one-third raised to the power of negative three all raised to the power of negative one.

Convert the sentence into numerical form.

$\rm \rightarrow \left [ \left (\dfrac{5}{7} \right )^2 \times \left ( \dfrac{1}{3}\right )^{-3} \right ] ^{-1}$

Simplify the expression, then we have

$\rm \rightarrow \left [ \left (\dfrac{7}{5} \right )^{2} \times \left ( \dfrac{1}{3}\right )^{3} \right ]$

The equivalent expression is seven-fifths squared times one-third cubed. Then the correct option is D.

More about the equivalent link is given below.

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2 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

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