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1 year ago

B. What are the second and third terms of this arithmetic sequence? 80, 페, 페, 125,...........

Mathematics

1 answer:

nataly862011 [7]1 year ago

8 0

The 2nd and 3rd term of an AP is found to be (a₂ = 95) and (a₃ = 110).

<h3>What is the sequence of AP arithmetic progression?</h3>

In Arithmetic Progression, the difference between the two numerical orders is a fixed number (AP). Arithmetic Sequence is another name for it.

We'd come across a few key concepts in AP that had been labeled as:

The first term (a)
Common difference (d)
Term nth (an)
The total of first n terms (Sn)

As shown below, the AP can also be referred to in terms of common differences.

The following is the procedure for evaluating an AP's n-th term: an = a + (n − 1) × d
The arithmetic progression sum is as follows: Sn = n/2[2a + (n − 1) × d].
Common difference 'd' of an AP: d = a2 - a1 = a3 - a2 = a4 - a3 = ...... = an - an-1.

Now, the given sequence is; 80, _, _, 125.

The series comprises of four given terms.

Let the first term be 'a₁' = 180.

The second term be 'a₂'.

The third term be 'a₃'.

And, the fourth term is 'a₄' = 125.

Use the nth term formula to find the common difference 'd'.

n-th term: an = a + (n − 1) × d

a₄ = a + (n - 1)d

125 = 80 + (4 - 1)d

45 = 3d

d = 15

Thus, the common difference is 15.

The second term is calculated as;

a₂ = a₁ + d

a₂ = 80 + 15

a₂ = 95.

The third term is estimated as;

a₃ = a₂ + d

a₃ = 95 = 15

a₃ = 110

Therefore, the 2nd and third term of an AP is computed as 95 and 110.

To know more about Arithmetic Sequence, here

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Find equations of the spheres with center(3, −4, 5) that touch the following planes.a. xy-plane b. yz- plane c. xz-plane

postnew [5]

Answer:

(a) (x - 3)² + (y + 4)² + (z - 5)² = 25

(b) (x - 3)² + (y + 4)² + (z - 5)² = 9

Step-by-step explanation:

The equation of a sphere is given by:

(x - x₀)² + (y - y₀)² + (z - z₀)² = r² ---------------(i)

Where;

(x₀, y₀, z₀) is the center of the sphere

r is the radius of the sphere

Given:

Sphere centered at (3, -4, 5)

=> (x₀, y₀, z₀) = (3, -4, 5)

(a) To get the equation of the sphere when it touches the xy-plane, we do the following:

i. Since the sphere touches the xy-plane, it means the z-component of its centre is 0.

Therefore, we have the sphere now centered at (3, -4, 0).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, -4, 0) as follows;

$d = \sqrt{(3-3)^2+ (-4 - (-4))^2 + (0-5)^2}$

$d = \sqrt{(3-3)^2+ (-4 + 4)^2 + (0-5)^2}$

$d = \sqrt{(0)^2+ (0)^2 + (-5)^2}$

$d = \sqrt{(25)}$

d = 5

This distance is the radius of the sphere at that point. i.e r = 5

Now substitute this value r = 5 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 5²

(x - 3)² + (y + 4)² + (z - 5)² = 25

Therefore, the equation of the sphere when it touches the xy plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 25

(b) To get the equation of the sphere when it touches the yz-plane, we do the following:

i. Since the sphere touches the yz-plane, it means the x-component of its centre is 0.

Therefore, we have the sphere now centered at (0, -4, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (0, -4, 5) as follows;

$d = \sqrt{(0-3)^2+ (-4 - (-4))^2 + (5-5)^2}$

$d = \sqrt{(-3)^2+ (-4 + 4)^2 + (5-5)^2}$

$d = \sqrt{(-3)^2 + (0)^2+ (0)^2}$

$d = \sqrt{(9)}$

d = 3

This distance is the radius of the sphere at that point. i.e r = 3

Now substitute this value r = 3 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 3²

(x - 3)² + (y + 4)² + (z - 5)² = 9

Therefore, the equation of the sphere when it touches the yz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 9

(b) To get the equation of the sphere when it touches the xz-plane, we do the following:

i. Since the sphere touches the xz-plane, it means the y-component of its centre is 0.

Therefore, we have the sphere now centered at (3, 0, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, 0, 5) as follows;

$d = \sqrt{(3-3)^2+ (0 - (-4))^2 + (5-5)^2}$

$d = \sqrt{(3-3)^2+ (0+4)^2 + (5-5)^2}$

$d = \sqrt{(0)^2 + (4)^2+ (0)^2}$

$d = \sqrt{(16)}$

d = 4

This distance is the radius of the sphere at that point. i.e r = 4

Now substitute this value r = 4 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 4²

(x - 3)² + (y + 4)² + (z - 5)² = 16

Therefore, the equation of the sphere when it touches the xz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 16

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