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2 years ago

Which sequences are arithmetic? select three options

Mathematics

2 answers:

Gemiola [76]2 years ago

6 0

Answer:

The sequences are arithmetic

1).-8.6, -5.0, -1.4, 2.2, 5.8

2). 5, 1, -3, -7, -11

3). -3, 3, 9, 15, 21

Step-by-step explanation:

If a sequence is a an AP then there is a common difference d.

1) Check sequence 1

-8.6, -5.0, -1.4, 2.2, 5.8

-5.0 - - 8.6 = 3.6

-1.4 - -5.4 = 3.6 It is an AP

2) Check sequence 2

2, -2.2,2.42, -2.662, 2.9282

-2.2 - 2 = -4.2

-2.662 - 2.42 = -5.082 Not AP

Similarly AP sequences are

5, 1, -3, -7, -11

-3, 3, 9, 15, 21

kaheart [24]2 years ago

6 0

Answer:

Options 1, 3 and 4.

Step-by-step explanation:

As we know in an arithmetic sequence each successive term has a common difference.

Let's check each option for Arithmetic sequence.

Option 1.

-8.6, -5.0, -1.4, 2.2, 5.8,....

Difference between 2nd and 1st term

-5 - (-8.6) = -5 + 8.6

= 3.6

Difference between 4th and 3rd term

-1.4 - (-5) = -1.4 + 5

= 3.6

This proves that there is a common difference of 3.6 which shows that the sequence is an arithmetic sequence.

Option 2.

2, -2.2, 2.42, -2.662, 2.9282,....

2nd term - 1st term = (-2.2) - 2 = -4.2

3rd term - 2nd term = 2.42 - (-2.2) = 2.42 + 2.2 = 4.62

Here difference in each term of the sequence is not common so it's not an arithmetic sequence.

Option 3.

5, 1, -3, -7, -11,.......

2nd term - 1st term = 1 - 5 = -4

3rd term - 2nd term = -3 - 1 = -4

Therefore, the given sequence is an arithmetic sequence.

Option 4.

-3, 3, 9, 15, 21,.......

2nd term - 1st term = 3 - (-3) = 3+ 3 = 6

3rd term - 2nd term = 9 - 3 = 6

There is a common difference of 6 in each term so it's an arithmetic sequence.

Option 5.

-6.2, -3.1, -1.55, -0.775, -0.3875,.........

2nd term - 1st term = -3.1 - (-6.2) = 6.2 - 3.1 = 3.1

3rd term - 2nd term = -1.55 - (-3.1) = 3.1 - 1.55 = 1.55

And 3.1 ≠ 1.55 which shows that this sequence is not an arithmetic sequence.

Options 1, 3 and 4 are arithmetic sequences.

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Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :

$E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1$

so we end up with

$E[Y^n]=e^{\frac12n^2}$

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2 years ago

Which sequences are arithmetic? select three options​

Which sequences are arithmetic? select three options