The problem describes the jumps of a toy rabbit. The lengths of the jumps form a sequence: 96 cm, 90 cm, and 84 cm. i. Show that the lengths of the jumps form an arithmetic progression. ii. Find the length of the 16th jump. iii. Based on (ii), how many jumps does the rabbit make before stopping (distance is 0)? iv. Find the total distance covered by the rabbit before stopping. v. The rabbit takes 20 seconds to stop. A toy turtle's speed is $40 \text{ cm/s}$. How far behind is the turtle when the rabbit stops?

1. Problem Description

The problem describes the jumps of a toy rabbit. The lengths of the jumps form a sequence: 96 cm, 90 cm, and 84 cm.

i. Show that the lengths of the jumps form an arithmetic progression.

ii. Find the length of the 16th jump.

iii. Based on (ii), how many jumps does the rabbit make before stopping (distance is 0)?

iv. Find the total distance covered by the rabbit before stopping.

v. The rabbit takes 20 seconds to stop. A toy turtle's speed is $40 \text{ cm/s}$. How far behind is the turtle when the rabbit stops?

2. Solution Steps

i. To show that the jump lengths form an arithmetic progression, we need to show that the difference between consecutive terms is constant.

Difference between the first two terms: $90 - 96 = -6$ cm.

Difference between the second and third terms: $84 - 90 = -6$ cm.

Since the common difference is constant ($d = -6$), the jump lengths form an arithmetic progression.

ii. We need to find the 16th term ($a_{16}$) of the arithmetic progression. The first term is $a_1 = 96$ cm, and the common difference is $d = -6$ cm. The formula for the nth term of an arithmetic progression is

$a_n = a_1 + (n-1)d$.

So, $a_{16} = 96 + (16-1)(-6) = 96 + 15(-6) = 96 - 90 = 6$ cm.

iii. We need to find the number of jumps (n) until the jump length is 0 or negative. We need to find $n$ such that $a_n \le 0$.

$a_n = a_1 + (n-1)d = 96 + (n-1)(-6) \le 0$

$96 - 6(n-1) \le 0$

$96 \le 6(n-1)$

$16 \le n-1$

$17 \le n$

Therefore, the number of jumps is

1

7.

iv. We need to find the sum of the arithmetic series up to the 16th term where all the terms are positive. We have calculated $a_{17} = 96 + (17-1)(-6) = 96 - 96 = 0$

The sum of the first n terms of an arithmetic series is given by $S_n = \frac{n}{2}(a_1 + a_n)$.

$S_{16} = \frac{16}{2}(96 + 6) = 8(102) = 816$ cm.

Or

$S_{17} = \frac{17}{2}(96+0) = \frac{17}{2} (96) = 17(48) = 816$ cm.

$S_{17} = 816$ cm

v. The turtle's speed is $40 \text{ cm/s}$. The time is 20 seconds. Distance covered by the turtle is given by:

Distance = Speed * Time

Distance = $40 \text{ cm/s} * 20 \text{ s} = 800 \text{ cm}$.

The rabbit has covered 816 cm.

The turtle is $816 - 800 = 16$ cm behind the rabbit.

3. Final Answer

i. The common difference is -6 cm, therefore it is an arithmetic progression.

ii. 6 cm

iii. 17

iv. 816 cm

v. 16 cm