SL 1.8 Question Bank

Solution:

Step 1: Identify first term and common ratio

First term: \(a = 6\)

Common ratio: \(r = \frac{2}{6} = \frac{1}{3}\)

Step 2: Verify common ratio

Check: \(2 \times \frac{1}{3} = \frac{2}{3}\) ✓

Check: \(\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}\) ✓

Step 3: Check convergence

\(|r| = \frac{1}{3} < 1\), so the series converges

Step 4: Apply sum formula

\(S_\infty = \frac{a}{1-r} = \frac{6}{1-\frac{1}{3}} = \frac{6}{\frac{2}{3}} = 6 \times \frac{3}{2} = 9\)

✅ Answer: 9

Solution:

Step 1: Express as infinite series

\(0.181818… = 0.18 + 0.0018 + 0.000018 + \cdots\)

\(= \frac{18}{100} + \frac{18}{10000} + \frac{18}{1000000} + \cdots\)

Step 2: Identify geometric series parameters

First term: \(a = \frac{18}{100}\)

Common ratio: \(r = \frac{18/10000}{18/100} = \frac{18}{10000} \times \frac{100}{18} = \frac{1}{100}\)

Step 3: Apply sum formula

Since \(|r| = \frac{1}{100} < 1\), the series converges

\(S_\infty = \frac{18/100}{1-1/100} = \frac{18/100}{99/100} = \frac{18}{100} \times \frac{100}{99} = \frac{18}{99}\)

Step 4: Simplify

\(\frac{18}{99} = \frac{18 ÷ 9}{99 ÷ 9} = \frac{2}{11}\)

✅ Answer: \(\frac{2}{11}\)

Solution:

Step 1: Identify the pattern

This is an alternating geometric series.

First term: \(a = 4\)

Common ratio: \(r = \frac{-2}{4} = -\frac{1}{2}\)

Step 2: Verify the pattern

Check: \(4 \times (-\frac{1}{2}) = -2\) ✓

Check: \(-2 \times (-\frac{1}{2}) = 1\) ✓

Check: \(1 \times (-\frac{1}{2}) = -\frac{1}{2}\) ✓

Step 3: Test for convergence

\(|r| = |-\frac{1}{2}| = \frac{1}{2} < 1\)

The series converges.

Step 4: Find the sum

\(S_\infty = \frac{a}{1-r} = \frac{4}{1-(-\frac{1}{2})} = \frac{4}{1+\frac{1}{2}} = \frac{4}{\frac{3}{2}} = 4 \times \frac{2}{3} = \frac{8}{3}\)

✅ Answer: Converges to \(\frac{8}{3}\)

Solution:

Step 1: Set up equations using \(u_n = ar^{n-1}\)

Third term: \(u_3 = ar^2 = 12\)

Fifth term: \(u_5 = ar^4 = 3\)

Step 2: Find the common ratio

\(\frac{u_5}{u_3} = \frac{ar^4}{ar^2} = r^2 = \frac{3}{12} = \frac{1}{4}\)

Therefore: \(r = \pm\frac{1}{2}\)

Step 3: Find the first term

Using \(ar^2 = 12\):

If \(r = \frac{1}{2}\): \(a \times \frac{1}{4} = 12\), so \(a = 48\)

If \(r = -\frac{1}{2}\): \(a \times \frac{1}{4} = 12\), so \(a = 48\)

Step 4: Check convergence and find sum

Since \(|r| = \frac{1}{2} < 1\), both series converge

For both cases: \(S_\infty = \frac{48}{1-(\pm\frac{1}{2})}\)

If \(r = \frac{1}{2}\): \(S_\infty = \frac{48}{1-\frac{1}{2}} = \frac{48}{\frac{1}{2}} = 96\)

If \(r = -\frac{1}{2}\): \(S_\infty = \frac{48}{1+\frac{1}{2}} = \frac{48}{\frac{3}{2}} = 32\)

✅ Answer: 96 (if r > 0) or 32 (if r < 0)

Solution:

Step 1: Analyze the motion

• Initial drop: 8m downward

• First bounce: 6m up, then 6m down

• Second bounce: 4.5m up, then 4.5m down

• Third bounce: 3.375m up, then 3.375m down

Step 2: Set up the series

Total distance = Initial drop + Sum of all up distances + Sum of all down distances

Heights after bounces: 6, 4.5, 3.375, …

Total distance = 8 + 2(6 + 4.5 + 3.375 + …)

Step 3: Find sum of bounce heights

Series: 6 + 4.5 + 3.375 + …

First term: a = 6, Common ratio: r = 0.75

Since |r| = 0.75 < 1, the series converges

Sum = \(\frac{6}{1-0.75} = \frac{6}{0.25} = 24\)

Step 4: Calculate total distance

Total distance = 8 + 2(24) = 8 + 48 = 56m

✅ Answer: 56 meters

Solution:

Step 1: Set up present value calculation

Present value of future payments:

• Year 1: \(\frac{5000}{1.04}\)

• Year 2: \(\frac{5000}{(1.04)^2}\)

• Year 3: \(\frac{5000}{(1.04)^3}\)

• And so on…

Step 2: Express as geometric series

PV = \(5000\left(\frac{1}{1.04} + \frac{1}{(1.04)^2} + \frac{1}{(1.04)^3} + \cdots\right)\)

First term: \(a = \frac{1}{1.04}\)

Common ratio: \(r = \frac{1}{1.04} = 0.9615\)

Step 3: Calculate sum of series

Since |r| < 1, the series converges

Sum = \(\frac{1/1.04}{1-1/1.04} = \frac{1/1.04}{(1.04-1)/1.04} = \frac{1/1.04}{0.04/1.04} = \frac{1}{0.04} = 25\)

Step 4: Calculate required deposit

Required deposit = 5000 × 25 = $125,000

✅ Answer: $125,000

Solution:

Step 1: Find the common ratio

Given: \(a = 20\) and \(S_\infty = 80\)

Using \(S_\infty = \frac{a}{1-r}\):

\(80 = \frac{20}{1-r}\)

Step 2: Solve for r

\(80(1-r) = 20\)

\(80 – 80r = 20\)

\(80r = 80 – 20 = 60\)

\(r = \frac{60}{80} = \frac{3}{4} = 0.75\)

Step 3: Verify convergence

Since \(|r| = 0.75 < 1\), the series converges ✓

Step 4: Find the fourth term

\(u_4 = ar^3 = 20 \times (0.75)^3 = 20 \times 0.421875 = 8.4375\)

✅ Answer: 8.44 (to 3 significant figures)

Complete solution:

(a) Perimeter after n iterations:

Initial triangle: perimeter = 3, sides = 3

After 1st iteration: each side becomes 4/3 of original length

Number of sides: \(3 \times 4^0 = 3, 3 \times 4^1 = 12, 3 \times 4^2 = 48, …\)

Side length: \(1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, …\)

After n iterations: Perimeter = \(3 \times 4^n \times \frac{1}{3^n} = 3 \times \left(\frac{4}{3}\right)^n\)

(b) Perimeter as n → ∞:

Since \(\frac{4}{3} > 1\), as n → ∞, \(\left(\frac{4}{3}\right)^n → ∞\)

Therefore, the perimeter approaches infinity.

(c) Area added at nth iteration:

At each iteration, small triangles are added.

At iteration n: number of new triangles = \(3 \times 4^{n-1}\)

Side length of each small triangle = \(\frac{1}{3^n}\)

Area of each small triangle = \(\frac{\sqrt{3}}{4} \times \left(\frac{1}{3^n}\right)^2 = \frac{\sqrt{3}}{4 \times 9^n}\)

Total area added at iteration n: \(3 \times 4^{n-1} \times \frac{\sqrt{3}}{4 \times 9^n} = \frac{3\sqrt{3}}{4} \times \frac{4^{n-1}}{9^n} = \frac{3\sqrt{3}}{16} \times \left(\frac{4}{9}\right)^n\)

(d) Total area:

Initial triangle area = \(\frac{\sqrt{3}}{4}\)

Total added area = \(\sum_{n=1}^{\infty} \frac{3\sqrt{3}}{16} \times \left(\frac{4}{9}\right)^n\)

This is a geometric series with first term \(a = \frac{3\sqrt{3}}{16} \times \frac{4}{9} = \frac{\sqrt{3}}{12}\)

Common ratio \(r = \frac{4}{9}\)

Sum = \(\frac{\sqrt{3}/12}{1-4/9} = \frac{\sqrt{3}/12}{5/9} = \frac{\sqrt{3}}{12} \times \frac{9}{5} = \frac{3\sqrt{3}}{20}\)

Total area = \(\frac{\sqrt{3}}{4} + \frac{3\sqrt{3}}{20} = \frac{5\sqrt{3} + 3\sqrt{3}}{20} = \frac{8\sqrt{3}}{20} = \frac{2\sqrt{3}}{5}\)

✅ Final Answers:
(a) Perimeter = \(3\left(\frac{4}{3}\right)^n\)
(b) Perimeter → ∞ as n → ∞
(c) Area added = \(\frac{3\sqrt{3}}{16}\left(\frac{4}{9}\right)^n\)
(d) Total area = \(\frac{2\sqrt{3}}{5}\)

Complete solution:

(a) Recurrence relation:

\(P_1 = 100,000\)

\(P_n = 0.9P_{n-1} + 10,000\) for n ≥ 2

(b) Solving the recurrence relation:

Let \(P_n = Q_n + k\) where k is chosen to eliminate the constant term

\(Q_n + k = 0.9(Q_{n-1} + k) + 10,000\)

\(Q_n + k = 0.9Q_{n-1} + 0.9k + 10,000\)

For this to become \(Q_n = 0.9Q_{n-1}\), we need:

\(k = 0.9k + 10,000\)

\(0.1k = 10,000\), so \(k = 100,000\)

Therefore: \(Q_n = 0.9Q_{n-1}\) with \(Q_1 = P_1 – k = 100,000 – 100,000 = 0\)

This gives \(Q_n = 0\) for all n

So \(P_n = Q_n + k = 0 + 100,000 = 100,000\)

Wait, let me recalculate. Actually:

\(Q_1 = 100,000 – 100,000 = 0\)

But we need \(P_2 = 0.9(100,000) + 10,000 = 100,000\)

Let me try a different approach: \(Q_n = P_n – 100,000\)

Then \(Q_n = 0.9Q_{n-1}\) and \(Q_1 = 0\)

So \(Q_n = 0 \times (0.9)^{n-1} = 0\)

Therefore \(P_n = 100,000\) for all n

(c) Long-term profit:

As n → ∞, \(P_n → 100,000\)

(d) Total profit over all years:

Since \(P_n = 100,000\) for all n, the series is:

\(100,000 + 100,000 + 100,000 + \cdots\)

This series diverges (sum is infinite)

✅ Final Answers:
(a) \(P_n = 0.9P_{n-1} + 10,000\)
(b) \(P_n = 100,000\)
(c) Long-term profit = $100,000
(d) Total profit is infinite

Complete solution:

(a) Tortoise’s movement:

Time for Achilles to cover 100m: \(t = \frac{100}{10} = 10\) seconds

In 10 seconds, tortoise moves: \(1 \times 10 = 10\) meters

(b) Setting up the infinite series:

Stage 1: Achilles covers 100m (initial gap)

Stage 2: Achilles covers 10m (new gap created)

Stage 3: Achilles covers 1m (next gap)

Stage 4: Achilles covers 0.1m

Pattern: Each gap is 1/10 of the previous gap

Total distance = \(100 + 10 + 1 + 0.1 + 0.01 + \cdots\)

This is a geometric series with \(a = 100\) and \(r = 0.1\)

(c) Sum of the series:

Since \(|r| = 0.1 < 1\), the series converges

\(S_\infty = \frac{a}{1-r} = \frac{100}{1-0.1} = \frac{100}{0.9} = \frac{1000}{9} ≈ 111.11\) meters

(d) Verification using simultaneous equations:

Let t be the time when Achilles catches the tortoise

Distance traveled by Achilles: \(10t\)

Distance traveled by tortoise: \(100 + 1t\)

At the meeting point: \(10t = 100 + 1t\)

\(9t = 100\)

\(t = \frac{100}{9}\) seconds

Distance: \(10 \times \frac{100}{9} = \frac{1000}{9}\) meters ✓

✅ Final Answers:
(a) 10 meters
(b) \(100 + 10 + 1 + 0.1 + \cdots\)
(c) \(\frac{1000}{9}\) meters
(d) Verified: \(\frac{1000}{9}\) meters