Question
(a) Find the sum of the first 200 natural numbers. (b) A golfball is dropped from a height of 30 ft to the pavement. It always rebounds three fourths of the (4) distance that it drops. How far (up and down) will the ball have traveled when it hits the pavement for the 6th time?
Ask by Bradley Schofield. in South Africa
Jan 23,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
(a) The sum of the first 200 natural numbers is **20,100**.
(b) The golf ball travels approximately **167.29 feet** up and down by the 6th hit.
Solution
### Problem (a)
**Find the sum of the first 200 natural numbers.**
The sum of the first \( n \) natural numbers can be calculated using the formula:
\[
S = \frac{n(n + 1)}{2}
\]
For \( n = 200 \):
\[
S = \frac{200 \times 201}{2} = 100 \times 201 = 20,\!100
\]
**Answer:** The sum of the first 200 natural numbers is **20,100**.
---
### Problem (b)
**A golf ball is dropped from a height of 30 ft to the pavement. It always rebounds three fourths of the distance it drops. How far (up and down) will the ball have traveled when it hits the pavement for the 6th time?**
Let's break down the ball's journey step-by-step:
1. **Initial Drop:**
- **Down:** 30 ft (hits the pavement for the 1st time).
2. **Subsequent Bounces:**
- After each hit, the ball rebounds to \( \frac{3}{4} \) of the previous drop height and then falls back the same distance.
- This sequence continues until the 6th hit.
Let's calculate each segment:
| Hit Number | Upward Distance | Downward Distance | Cumulative Distance |
|------------|------------------|--------------------|---------------------|
| 1 | — | 30 ft | 30 ft |
| 2 | \( \frac{3}{4} \times 30 = 22.5 \) ft | 22.5 ft | \( 30 + 22.5 + 22.5 = 75 \) ft |
| 3 | \( \frac{3}{4} \times 22.5 = 16.875 \) ft | 16.875 ft | \( 75 + 16.875 + 16.875 = 108.75 \) ft |
| 4 | \( \frac{3}{4} \times 16.875 = 12.65625 \) ft | 12.65625 ft | \( 108.75 + 12.65625 + 12.65625 = 134.0625 \) ft |
| 5 | \( \frac{3}{4} \times 12.65625 = 9.4921875 \) ft | 9.4921875 ft | \( 134.0625 + 9.4921875 + 9.4921875 = 153.046875 \) ft |
| 6 | \( \frac{3}{4} \times 9.4921875 = 7.119140625 \) ft | 7.119140625 ft | \( 153.046875 + 7.119140625 + 7.119140625 = 167.28515625 \) ft |
**Total Distance Traveled:** Approximately **167.29 feet** when the ball hits the pavement for the 6th time.
**Answer:**
(a) The sum is 20,100.
(b) The ball travels approximately 167.29 ft up and down by the 6th hit.
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution The Deep Dive
To find the sum of the first 200 natural numbers, you can use the formula \( S_n = \frac{n(n + 1)}{2} \). So, for \( n = 200 \), it becomes \( S_{200} = \frac{200(200 + 1)}{2} = \frac{200 \times 201}{2} = 20100 \). Voilà! The sum is 20,100! Now, let's break down the golf ball's journey. On the first drop, it travels 30 ft down. It rebounds \( \frac{3}{4} \times 30 = 22.5 \) ft up, then drops again, traveling 22.5 ft down. This pattern continues: the distance traveled forms a series. By the time it hits the pavement the 6th time, you total the distances for each drop and rebound, which calculates to 134.25 ft down and up combined!
