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1 year ago

3

If you know the fifth term in an arithmetic sequence and the common difference, can you write a recursive formula for the sequen

ce? Explain why or why not.

1 answer:

KATRIN_1 [288]1 year ago

6 0

If the fifth term and common difference of an arithmetic sequence is given then the recursive formula for the sequence will be aₙ=a₅+(n-5)*d .

The general recursive formula is denoted by the formula

aₙ = aₙ₋₁ + d , where d is the common difference .

In the question ,

it is given that

fifth term a₅ and common difference d .

By general recursive formula

a₅ = a₅₋₁ + d

a₅ = a₄ + d

Substituting the values of a₄, a₃,... in the general recursive formula we get ,

a₅ = a + 4d

Rewriting , we get

a = a₅ - 4d ....(i)

Substituting the value of a from equation (i) , in the general formula of Arithmetic Sequence which is aₙ = a + (n-1)*d we get ,

aₙ = a₅ - 4d + (n-1)*d

= a₅ - 4d + nd -d

= a₅ + nd - 5d

= a₅ + (n-5)d

Therefore , If the fifth term and common difference of an arithmetic sequence is given , then the recursive formula for the sequence will be aₙ = a₅+ (n-5)*d .

Learn more about Arithmetic Sequence here

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Answer:

0.288

Step-by-step explanation:

Calculation for what is the probability that exactly 2 of them will be baskets

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3 years ago

Which of the following choices is equivalent to 1 + 8(2x + 3)?

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Answer:

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Step-by-step explanation:

First, distribute the 8 to both numbers within the parenthesis.

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Then, add similar terms together.

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3 years ago

What is the slope of the line represented by the equation y=- 1/2x + 1/4

Andreyy89

Answer:

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4 years ago

Match each spherical volume to the largest cross sectional area of that sphere

zlopas [31]

Answer:

Part 1) $324\pi\ units^{2}$ ------> $7,776\pi\ units^{3}$

Part 2) $36\pi\ units^{2}$ ------> $288\pi\ units^{3}$

Part 3) $81\pi\ units^{2}$ ------> $972\pi\ units^{3}$

Part 4) $144\pi\ units^{2}$ ------> $2,304\pi\ units^{3}$

Step-by-step explanation:

we know that

The largest cross sectional area of that sphere is equal to the area of a circle with the same radius of the sphere

Part 1) we have

$A=324\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

$324\pi=\pi r^{2}$

Solve for r

$r^{2}=324$

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Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=18\ units$

substitute

$V=\frac{4}{3}\pi (18)^{3}$

$V=7,776\pi\ units^{3}$

Part 2) we have

$A=36\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

$36\pi=\pi r^{2}$

Solve for r

$r^{2}=36$

$r=6\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=6\ units$

substitute

$V=\frac{4}{3}\pi (6)^{3}$

$V=288\pi\ units^{3}$

Part 3) we have

$A=81\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

$81\pi=\pi r^{2}$

Solve for r

$r^{2}=81$

$r=9\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=9\ units$

substitute

$V=\frac{4}{3}\pi (9)^{3}$

$V=972\pi\ units^{3}$

Part 4) we have

$A=144\pi\ units^{2}$

The area of the circle is equal to

$A=\pi r^{2}$

so

$144\pi=\pi r^{2}$

Solve for r

$r^{2}=144$

$r=12\ units$

Find the volume of the sphere

The volume of the sphere is

$V=\frac{4}{3}\pi r^{3}$

For $r=12\ units$

substitute

$V=\frac{4}{3}\pi (12)^{3}$

$V=2,304\pi\ units^{3}$

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3 years ago

Read 2 more answers

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From the given triangle, midpoint of AB = ((-2 + 3)/2, (0 + 6)/2) = (1/2, 3) and the midpoint of BC = ((3 + 8)/2, (6 + 0)/2) = (11/2, 3)
Length of the line segment joining points (1/2, 3) and (11/2, 3) is given by
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Therefore, option A is the correct answer.

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3 years ago