Question

How do I solve these problems?

Answer 1

Answer:

-32, -7, 18 and

127, 106, 85

Step-by-step explanation:

You can use the following formula to solve this:

$a_{n} = a_{1} + (n - 1)d$

a is the nth term in the sequence. So at the 3rd term, n = 3

a₁ is the first term in the sequence. In your first question, a₁ is -32. In your second, a₁ is 127.

(n - 1) represents the term before a, or the term before the current term. So at the 3rd term, (n - 1) = 3-1, which is 2.

d represents the common difference between terms. Aka what you add or subtract to a term to get the next one.

For the first question, you know a₁ and d, so you can substitute them into the equation to find the second and third terms:

a = -32 + (n - 1)25

You also know (n-1), because a is 2 when you're looking for the second term. So (n-1) = 2-1, which is 1. So put that in the formula and simplify:

a = -32 + (1)25

a = -32 + 25

a = -7

The second term is -7. To find the 3rd term, do the same thing.

a = -32 + (n - 1)25

(n-1) is 2, since you're on the third term and n is 3.

a = -32 + (2)25

a = -32 + 50

a = 18

So the first 3 terms are -32, -7, and 18.

For the second one, the formula looks like this:

a = 127 + (n - 1)-21

The first term is 127. Find the 2nd (n is 2):

a = 127 + (1)-21

a = 127 -21

a = 106

Then find the 3rd term (n is 3)

a = 127 + (2)-21

a = 127 -42

a = 85

So the first 3 terms are 127, 106, and 85.

Answer 2

Answer:

- 32, - 7, 18 and 127, 106, 85

Step-by-step explanation:

Add the common difference d to the previous term

(1)

a₁ = - 32

a₂ = a₁ + 25 = - 32 + 25 = - 7

a₃ = a₂ + 25 = - 7 + 25 = 18

The first 3 terms are - 32, - 7, 18

(2)

a₁ = 127

a₂ = a₁ + d = 127 - 21 = 106

a₃ = a₂ + d = 106 - 21 = 85

The first 3 terms are 127, 106, 85