To solve the problem, we need to understand the snail's progress each day. During the day, the snail climbs up 3 meters but at night it slips back 2 meters.
Let's break down its daily progress:
- During the day, the snail climbs up to a height of \(3\) meters.
- At night, it slips back \(2\) meters.
Thus, the net gain per day is: \[ 3 \text{ meters (climbed)} - 2 \text{ meters (slipped)} = 1 \text{ meter per day}. \]
However, we need to consider that once the snail reaches the top of the well (20 meters), it will not slip back.
Now we'll calculate how many days it takes for the snail to reach or exceed 20 meters.
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Day 1:
- Climbs to \(3\) meters, slips back to \(1\) meter.
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Day 2:
- Climbs to \(4\) meters, slips back to \(2\) meters.
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Day 3:
- Climbs to \(5\) meters, slips back to \(3\) meters.
Continuing this process, we can summarize the position of the snail at the end of each day:
- After Day 1: 1 meter
- After Day 2: 2 meters
- After Day 3: 3 meters
- ...
- After Day 16: 16 meters
- After Day 17: 17 meters
- After Day 18: 18 meters
- After Day 19: 19 meters
Now, on Day 20, the snail starts at 19 meters:
- Climbs to 22 meters (3 meters up), which is above the well's height of 20 meters.
Therefore, the snail reaches the top of the well on the 20th day.
Thus, it will take the snail a total of 20 days to reach the top of the well.