8. A ball rolls down an inclined track and gains speed. If the distance the ball travels is 3 cm in the first second, 6 cm in the second, 9 cm in the third, and so on, find the total distance travelled in 15 seconds. ( 3 marks) 9. A superball bounces to \( \frac{3}{4} \) of its initial height when dropped on pavement. If the ball is dropped from a height of 16 m , how far has the ball travelled by the!time it hits the (4 marks) (4) for the \( 6^{\text {th }} \) time?

The pattern of the ball's distance traveled can be seen as an arithmetic sequence where the distance increases by 3 cm every second. To find the total distance traveled in 15 seconds, we can sum the distances of each second. The formula for the sum of an arithmetic series is \( S_n = \frac{n}{2} (a + l) \), where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term. Here, \( n = 15 \), \( a = 3 \), and \( l = 3 + 3(15 - 1) = 45 \). Thus, \( S_{15} = \frac{15}{2} (3 + 45) = \frac{15}{2} (48) = 360 \) cm. Now, for the superball, the distance traveled includes the drop and the subsequent bounces. The ball drops 16 m initially, and then for each bounce, it ascends to \( \frac{3}{4} \) of the previous height. The total distance when it hits the ground for the \( 6^{th} \) time will comprise the descending and ascending distances. The heights will be 16 m, \( 16 \times \frac{3}{4} \), \( 16 \times \left(\frac{3}{4}\right)^2 \), and so on, for a total of 6 bounces. This gives a distance of \( 16 + 2 \times \sum_{k=0}^{5} 16 \left(\frac{3}{4}\right)^k \). Using the geometric series formula \( S_n = a \frac{1 - r^n}{1 - r} \) for the heights of the bounces, the final distance can be calculated as \( 16 + 2 \times 16 \frac{1 - \left(\frac{3}{4}\right)^6}{1 - \frac{3}{4}} \).