Practice problems involving recursive sequences, including finding values and completing expressions.

Let's solve each part of the problem step by step.

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Problem 1:

A sequence is defined by:
- $ f(0) = -20 $
- $ f(n) = f(n-1) - 5 $ for $ n \geq 1 $

#### a. Explain why $ f(1) = -20 - 5 $.

We are given that:
- $ f(0) = -20 $
- The recursive rule: $ f(n) = f(n-1) - 5 $ for $ n \geq 1 $

To find $ f(1) $, plug in $ n = 1 $ into the recursive formula:
$$
f(1) = f(0) - 5
$$
Since $ f(0) = -20 $, we get:
$$
f(1) = -20 - 5
$$

✔ So, $ f(1) = -20 - 5 $ because the function subtracts 5 from the previous term, and the previous term ($ f(0) $) is $-20$.

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#### b. Explain why $ f(3) = -20 - 5 - 5 - 5 $.

Let’s compute $ f(3) $ step by step using the recursive definition:

- $ f(0) = -20 $
- $ f(1) = f(0) - 5 = -20 - 5 = -25 $
- $ f(2) = f(1) - 5 = -25 - 5 = -30 $
- $ f(3) = f(2) - 5 = -30 - 5 = -35 $

Now notice:
$$
f(3) = -20 - 5 - 5 - 5 = -35
$$

Each time we go to the next term, we subtract 5. From $ f(0) $ to $ f(3) $, we apply the rule three times, so we subtract 5 three times.

✔ Therefore, $ f(3) = -20 - 5 - 5 - 5 $ because we subtract 5 for each step from $ n = 1 $ to $ n = 3 $, which is 3 subtractions.

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#### c. Complete the expression: $ f(10) = -20 - \underline{\quad} $. Explain your reasoning.

We want to express $ f(10) $ in terms of $ -20 $ minus some amount.

From the pattern:
- $ f(1) = f(0) - 5 $
- $ f(2) = f(1) - 5 = f(0) - 5 - 5 $
- $ f(3) = f(2) - 5 = f(0) - 5 - 5 - 5 $
- ...
- So, $ f(n) = f(0) - 5n $

Therefore,
$$
f(10) = f(0) - 5 \times 10 = -20 - 50
$$

So the completed expression is:
$$
f(10) = -20 - \boxed{50}
$$

✔ Reasoning: Since each term decreases by 5, and we go from $ f(0) $ to $ f(10) $, we subtract 5 a total of 10 times. So we subtract $ 5 \times 10 = 50 $.

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Problem 2:

A sequence is defined by:
- $ f(0) = -4 $
- $ f(n) = f(n-1) - 2 $ for $ n \geq 1 $

We are to write a definition for the $ n^\text{th} $ term of the sequence.

This is an arithmetic sequence where:
- First term: $ f(0) = -4 $
- Common difference: $ d = -2 $

The general formula for the $ n^\text{th} $ term of an arithmetic sequence is:
$$
f(n) = f(0) + n \cdot d
$$

Substitute:
$$
f(n) = -4 + n(-2) = -4 - 2n
$$

✔ So, the explicit definition for the $ n^\text{th} $ term is:
$$
\boxed{f(n) = -4 - 2n}
$$

Alternatively, you can write it as:
$$
f(n) = -2n - 4
$$

Both are equivalent.

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✔ Final Answers:

#### Problem 1:
a. $ f(1) = -20 - 5 $ because $ f(1) = f(0) - 5 $ and $ f(0) = -20 $.
b. $ f(3) = -20 - 5 - 5 - 5 $ because we subtract 5 three times from $ f(0) $ to reach $ f(3) $.
c. $ f(10) = -20 - \boxed{50} $, since we subtract 5 ten times.

#### Problem 2:
The $ n^\text{th} $ term is:
$$
\boxed{f(n) = -4 - 2n}
$$