Question

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

197

189

1805

1797

Answer 1

`a_(101)=197`

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.

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`