Question

What is the 101st term in the sequence 997, 989, 981, ...?

Answer 1

Answer:

$a_{101}=197$

Step-by-step explanation:

The first term of the sequence is

$a_1=997$

The given sequence  is 997, 989, 981, ...

The common difference is

$d=989-997=-8$

The nth ter of the sequence is

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

We plug in the first term and the common ratio to obtain;

$a_n=997-8(n-1)$

$a_n=997-8n+8$

$a_n=1005-8n$

We substitute n=101 to get;

$a_{101}=1005-8(101)$

$a_{101}=197$

Answer 2

Answer: first option.

Step-by-step explanation:

Find the common difference d of the arithmetic sequence:

$d=a_n-a_{(n-1)}\\d=989-997\\d=-8$

Then the formula for the 101st term is the shown below:

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

Where:

$a_1=997\\d=-8\\n=101$

Substitute values into the formula. Therefore, you obtain:

$a_n=997+(101-1)(-8)=197$