All categories

2 years ago

How do you determine the common difference for an arithmetic sequence?

Mathematics

1 answer:

antiseptic1488 [7]2 years ago

4 0

Answer:

Common difference(d) is:

$a2-a1=a3-a2=a4-a3= d$

d is constant till

$a_n-a_n_-_1 =d$

Step-by-step explanation:

An arithmetic sequence or arithmetic progression is the sequence, whose difference of consecutive terms called common difference is constant throughout the sequence.

Considering, the sequence $a_1,a_2,a_3,a_4,.......a_n$

$a2-a1=a3-a2=a4-a3= d$

d= common difference.

we can find any $n^t^h$ term in the sequence by using the formula

$a_n=a_1+(n-1)d$

where $a_n$ = any $n^t^h$ term

$n\\$ =Number of terms

$d$ =common difference

$a_1$ = first term

Also sum of n terms of A.P can be find by the formula:

$S_n=\frac{n}{2} [2a+(n-1)d]$

You might be interested in

Every week Roseanne gets paid $100 plus a 5% commission on all of her sales. If she has $5,000 in sales, how much is her pay for

EastWind [94]

Answer:

$700 a 30%

Step-by-step explanation:

hope it help :)

6 0

2 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span> $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

7 0

3 years ago

Tony’s club is selling oranges to raise money. For every box they sell, they get 1 1/8 dollars profit. They have sold 75 boxes a

dmitriy555 [2]

Tony’s club is selling oranges to raise money. For every box they sell, they get 1 1/8 dollars profit. They have sold 75 boxes already. How many more boxes must they sell to raise 180 dollars?

Answer: We are given:

The amount of money Tony's club get for every box they sell $=1 \frac{1}{8} =1.125$ dollar

They amount of money Tony's club has raised by selling 75 boxes is:

$75 \times 1.125=84.375$ dollars

The amount of money Tony's club is required to raise = 180 dollars

The remaining need to be raised is :

$180-84.375=95.625$ dollars

Therefore, the number of more boxes to be sold are:

$\frac{95.625}{1.125}=85$

Hence, 85 more boxes they must sell to raise 180 dollars

8 0

2 years ago

Ciara is naking a new outfit she needs 2 1/2 yards of the fabric for the shawl and 1 3/4 yards of fabric for the dress. If she h

valentina_108 [34]

Answer:

1 1/4 yards

$20

Step-by-step explanation:

The amount of fabric more that she needs can be determined by subtracting the sum of the yards needed for the shawl and the dress from the fabric she has

sum of the yards needed for the shawl and the dress =

$2\frac{1}{2} + 1\frac{3}{4}$