How to find the next term in an arithmetic sequence - PSAT Math

The series is defined by n2 – 1 starting at n = 1. The sixth number in the series then equal to 62 – 1 = 35.

Each term in the sequence is one less than twice the previous term.

So,

Notice that in the sequence

each term increases by .

It is always good strategy when attempting to find a pattern in a sequence to examine the difference between each term.

We continue the pattern to find:

The term is

The term is

The term is

It is useful to note that the sequence is defined by,

where n is the number of any one term.

We can solve

to find the term.

The pattern of the sequence is (x+1) * 2.

We have the first 5 terms, so we need terms 6 and 7:

(78+1) * 2 = 158

(158+1) * 2 = 318

3 + 8 + 18 +38 + 78 + 158 + 318 = 621

The series is defined by n2 – 1 starting at n = 1. The sixth number in the series then equal to 62 – 1 = 35.

Each term in the sequence is one less than twice the previous term.

So,

Notice that in the sequence

each term increases by .

It is always good strategy when attempting to find a pattern in a sequence to examine the difference between each term.

We continue the pattern to find:

The term is

The term is

The term is

It is useful to note that the sequence is defined by,

where n is the number of any one term.

We can solve

to find the term.

The pattern of the sequence is (x+1) * 2.

We have the first 5 terms, so we need terms 6 and 7:

(78+1) * 2 = 158

(158+1) * 2 = 318

3 + 8 + 18 +38 + 78 + 158 + 318 = 621

The series is defined by n2 – 1 starting at n = 1. The sixth number in the series then equal to 62 – 1 = 35.

Each term in the sequence is one less than twice the previous term.

So,

Notice that in the sequence

each term increases by .

It is always good strategy when attempting to find a pattern in a sequence to examine the difference between each term.

We continue the pattern to find:

The term is

The term is

The term is

It is useful to note that the sequence is defined by,

where n is the number of any one term.

We can solve

to find the term.

The pattern of the sequence is (x+1) * 2.

We have the first 5 terms, so we need terms 6 and 7:

(78+1) * 2 = 158

(158+1) * 2 = 318

3 + 8 + 18 +38 + 78 + 158 + 318 = 621

The series is defined by n2 – 1 starting at n = 1. The sixth number in the series then equal to 62 – 1 = 35.

Each term in the sequence is one less than twice the previous term.

So,

Notice that in the sequence

each term increases by .

It is always good strategy when attempting to find a pattern in a sequence to examine the difference between each term.

We continue the pattern to find:

The term is

The term is

The term is

It is useful to note that the sequence is defined by,

where n is the number of any one term.

We can solve

to find the term.

The pattern of the sequence is (x+1) * 2.

We have the first 5 terms, so we need terms 6 and 7:

(78+1) * 2 = 158

(158+1) * 2 = 318

3 + 8 + 18 +38 + 78 + 158 + 318 = 621

The series is defined by n2 – 1 starting at n = 1. The sixth number in the series then equal to 62 – 1 = 35.

Each term in the sequence is one less than twice the previous term.

So,

Notice that in the sequence

each term increases by .

It is always good strategy when attempting to find a pattern in a sequence to examine the difference between each term.

We continue the pattern to find:

The term is

The term is

The term is

It is useful to note that the sequence is defined by,

where n is the number of any one term.

We can solve

to find the term.