Which sequence could be partially defined by the recursive formula f (n + 1) = f (n) + 2.5 for n 2 1?

2.5, 6.25, 15.625, 39.0625, ...

2.5, 5, 10, 20

-10,-7.5, - 5, - 2.5, ...

-10, - 25, 62.5, 156.25

Sequence 3

-10,-7.5,-5,-2.5, ...

Step-by-step explanation:

So f (n+1) = f (n) + 2.5 means a term can be found by adding it's previous term to 2.5. That means this is an arithmetic sequence with a common difference of 2.5.

f (n+1) = f (n) + d is the recursive form for an arithmetic sequence with common difference d.

So you are looking for a sequence of numbers that is going up by 2.5 each time.

Let's check sequence 1:

2.5+2.5=5 so not this one because we didn't get 6.25 next.

Let's check sequence 2:

2.5+2.5=5 is what we have for the 2nd term.

5+2.5=7.5 so not this one because we didn't get 10 next.

Let's check sequence 3:

-10+2.5=-7.5 is the 2nd term

-7.5+2.5=-5 is the 3rd term

-5+2.5=-2.5 is the 4th term

Sequence 3 is arithmetic with common difference 2.5 assuming the pattern continues.

Let's check sequence 4 for fun:

-10+2.5=-7.5 is not - 25

So we are done. Sequence 3 is the only one that fits term=previous term+2.5 or f (n+1) = f (n) + 2.5.