Question

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

Answer 1

Option C: –10, –7.5, –5, –2.5, … is the sequence that could be partially defined by the recursive formula $f(n+1)=f(n)+2.5$

Explanation:

The given recursive formula is $f(n+1)=f(n)+2.5$ for $n\geq 1$ and $f(1)=-10$

We need to determine the sequence.

The sequence can be determined by substituting n = 1, 2, 3, 4,....

2nd term of the sequence:

Substituting n = 1 in the formula $f(n+1)=f(n)+2.5$, we get,

$f(1+1)=f(1)+2.5$

Simplifying, we have,

$f(2)=-10+2.5=-7.5$

Thus, the 2nd term of the sequence is -7.5

3rd term of the sequence:

Substituting n = 2 in the formula $f(n+1)=f(n)+2.5$, we get,

$f(2+1)=f(2)+2.5$

Simplifying, we have,

$f(3)=-7.5+2.5=-5$

Thus, the 3rd term of the sequence is -5

4th term of the sequence:

Substituting n = 3 in the formula $f(n+1)=f(n)+2.5$, we get,

$f(3+1)=f(3)+2.5$

Simplifying, we have,

$f(4)=-5+2.5=-2.5$

Thus, the 4th term of the sequence is -2.5

Therefore, the sequence is –10, –7.5, –5, –2.5, …

Hence, Option C is the correct answer.