IB Mathematics SL 1.3 Geometric sequences and series AA SL Paper 2- Exam Style Questions- New Syllabus

(a)
Heights form a geometric sequence with first term \( u_1 = 10 \) and common ratio \( r = \frac{2}{3} \).
Height after the 5^th bounce is the 6^th term of the sequence:
\( u_6 = 10 \times \left(\frac{2}{3}\right)^{5} \)
\( u_6 = 10 \times \frac{32}{243} = \frac{320}{243} \approx 1.31687… \)
\( \boxed{1.32 \text{ m}} \) (accept \( \frac{320}{243} \) m)

(b)
Distance before the 5^th bounce = initial drop + 4 complete up‑and‑down cycles.
Sum of the first 5 terms of the geometric sequence (initial drop plus four bounces):
\( S_5 = 10 + 10\left(\frac{2}{3}\right) + 10\left(\frac{2}{3}\right)^2 + 10\left(\frac{2}{3}\right)^3 + 10\left(\frac{2}{3}\right)^4 \)
Using \( S_n = \frac{u_1(1-r^n)}{1-r} \):
\( S_5 = \frac{10\left[1-\left(\frac{2}{3}\right)^5\right]}{1-\frac{2}{3}} = \frac{10\left(1-\frac{32}{243}\right)}{\frac{1}{3}} = 30\left(\frac{211}{243}\right) \approx 26.049… \)
Total distance travelled = initial drop + twice the sum of the four upward bounces:
\( \text{Distance} = 10 + 2 \times \left[10\left(\frac{2}{3}\right) + 10\left(\frac{2}{3}\right)^2 + 10\left(\frac{2}{3}\right)^3 + 10\left(\frac{2}{3}\right)^4\right] \)
Which simplifies to:
\( \text{Distance} = 10 + 2(S_5 – 10) = 2S_5 – 10 \approx 42.0987… \)
\( \boxed{42.1 \text{ m}} \)

(c)
For a drop from height \( x \), the total distance is the sum of the infinite geometric series of up‑and‑down movements.
Downward distances: \( x + xr + xr^2 + \dots \) with \( r = \frac{2}{3} \).
Upward distances: \( xr + xr^2 + xr^3 + \dots \).
Total:
\( \delta = x + 2\left( xr + xr^2 + xr^3 + \dots \right) \)
The infinite sum \( xr + xr^2 + \dots = \frac{xr}{1-r} \).
Thus:
\( \delta = x + 2 \cdot \frac{xr}{1-r} = x + 2 \cdot \frac{x \cdot \frac{2}{3}}{1-\frac{2}{3}} = x + 2 \cdot \frac{2x/3}{1/3} = x + 2 \cdot 2x = 5x \)
\( \boxed{\delta = 5x} \)

(d)
Using the result from (c) with \( x = 10 \):
\( \delta_1 = 5 \times 10 = 50 \)
\( \boxed{50 \text{ m}} \)

(e)
For Ball 2, \( x = 9.56 \), so
\( \delta_2 = 5 \times 9.56 = 47.8 \)
The sequence \( \delta_1, \delta_2, \delta_3, \dots \) is arithmetic with first term \( \delta_1 = 50 \) and common difference
\( d = \delta_2 – \delta_1 = 47.8 – 50 = -2.2 \)
The \( n \)-th term is
\( \delta_n = 50 + (n-1)(-2.2) = 50 – 2.2(n-1) \)
Set \( \delta_n < 25 \):
\( 50 – 2.2(n-1) < 25 \)
\( -2.2(n-1) < -25 \)
\( n-1 > \frac{25}{2.2} \approx 11.3636… \)
\( n > 12.3636… \)
Thus the first integer satisfying the inequality is \( n = 13 \).
\( \boxed{\text{Ball 13}} \)