The Common Ratio & General $n$th Term Formula
- Definition of a Geometric Sequence: A sequence in which each term after the first is found by multiplying the preceding term by a non-zero constant $r \in \mathbb{R} \setminus \{0\}$ called the common ratio: $$r = \frac{u_{k+1}}{u_k} = \frac{u_2}{u_1} = \frac{u_3}{u_2}$$
- The General $n$th Term Formula: $$u_n = u_1 r^{n-1}$$ where $u_1$ is the first term, $r$ is the common ratio, and $n \in \mathbb{Z}^+$ is the term position.
- Sign Behaviour:
- If $r > 0$, all terms have the same sign as $u_1$.
- If $r < 0$, the sequence is alternating (signs alternate between positive and negative due to $(-1)^{n-1}$).
- The Geometric Mean: Three numbers $a, b, c$ form consecutive terms of a geometric progression if and only if: $$\frac{b}{a} = \frac{c}{b} \implies b^2 = ac \implies b = \pm \sqrt{ac}$$
Example 1
Finding the Common Ratio & General Term
A geometric sequence begins $162, -108, 72, -48, \dots$
(a) Find the common ratio $r$ as an exact simplified fraction.
(b) Calculate the $7$th term $u_7$.
- Part (a): Compute the ratio of consecutive terms: $$r = \frac{u_2}{u_1} = \frac{-108}{162} = -\frac{2}{3}$$ Check with the next ratio: $\frac{u_3}{u_2} = \frac{72}{-108} = -\frac{2}{3}$.
- Part (b): Apply the $n$th term formula $u_n = u_1 r^{n-1}$: $$u_7 = 162 \left(-\frac{2}{3}\right)^{7 - 1} = 162 \left(-\frac{2}{3}\right)^6$$
- Evaluate the power: $$\left(-\frac{2}{3}\right)^6 = \frac{2^6}{3^6} = \frac{64}{729}$$ $$u_7 = 162 \times \frac{64}{729} = \frac{162 \times 64}{729}$$ Since $162 = 2 \times 81$ and $729 = 9 \times 81$: $$u_7 = \frac{2 \times 64}{9} = \frac{128}{9}$$
Answer: (a) $r = -\frac{2}{3}$; (b) $u_7 = \frac{128}{9}$
Solving Geometric Systems by Division
- System Architecture: When two non-consecutive terms $u_p$ and $u_q$ are given, set up two equations and divide them to cancel $u_1$: $$\frac{u_q}{u_p} = \frac{u_1 r^{q-1}}{u_1 r^{p-1}} = r^{q-p}$$
- Even vs. Odd Powers Caution:
- If $q - p$ is odd, there is exactly one real root: $r = \sqrt[q-p]{\frac{u_q}{u_p}}$.
- If $q - p$ is even, there are two possible real roots: $r = \pm \sqrt[q-p]{\frac{u_q}{u_p}}$. You must check if the question specifies "all positive terms" before discarding the negative root!
Example 2
Finding Parameters from Two Non-Consecutive Terms
In a geometric sequence with all positive terms, the $3$rd term is $24$ and the $7$th term is $384$. Find the common ratio $r$, the first term $u_1$, and the $10$th term $u_{10}$.
- Write expressions for $u_3$ and $u_7$: $$u_3 = u_1 r^2 = 24 \quad \text{--- (1)}$$ $$u_7 = u_1 r^6 = 384 \quad \text{--- (2)}$$
- Divide equation (2) by equation (1): $$\frac{u_1 r^6}{u_1 r^2} = \frac{384}{24} \implies r^4 = 16$$
- Solve for $r$: $$r = \pm \sqrt[4]{16} = \pm 2$$ Since the sequence has all positive terms, $r > 0 \implies r = 2$.
- Substitute $r = 2$ into (1) to find $u_1$: $$u_1 (2)^2 = 24 \implies 4u_1 = 24 \implies u_1 = 6$$
- Calculate $u_{10}$: $$u_{10} = 6 \times 2^{10 - 1} = 6 \times 2^9 = 6 \times 512 = 3072$$
Answer: $r = 2$, $u_1 = 6$, and $u_{10} = 3072$
Finite Geometric Series & Solving for $n$ via Logarithms
- Sum of First $n$ Terms ($S_n$): $$S_n = \frac{u_1(1 - r^n)}{1 - r} = \frac{u_1(r^n - 1)}{r - 1} \quad (r \neq 1)$$ Use the first form when $|r| < 1$, and the second form when $r > 1$ to keep the denominator positive.
- Solving for the Number of Terms $n$: When asked "how many terms are needed for the sum to exceed $K$," solve the inequality using logarithms: $$S_n > K \implies \frac{u_1(r^n - 1)}{r - 1} > K \implies r^n > 1 + \frac{K(r - 1)}{u_1}$$ Take the natural logarithm ($\ln$) of both sides: $$n \ln r > \ln\left(1 + \frac{K(r - 1)}{u_1}\right) \implies n > \frac{\ln(\dots)}{\ln r}$$ Remember: if taking logarithms where base < 1, $\ln r$ is negative, so flip the inequality sign!
Example 3
Solving for $n$ Using Logarithms
A geometric sequence has $u_1 = 3$ and $r = 2$. Find the smallest value of $n$ such that the sum of the first $n$ terms exceeds $10\,000$.
- Set up the inequality $S_n > 10\,000$: $$S_n = \frac{3(2^n - 1)}{2 - 1} = 3(2^n - 1) > 10\,000$$
- Isolate the exponential term $2^n$: $$2^n - 1 > \frac{10\,000}{3} \approx 3333.33$$ $$2^n > 3334.33$$
- Apply natural logarithms to both sides: $$\ln(2^n) > \ln(3334.33)$$ $$n \ln 2 > \ln(3334.33) \implies n > \frac{\ln(3334.33)}{\ln 2} \approx \frac{8.112}{0.693} \approx 11.70$$
- State the smallest integer $n$:
Since $n \in \mathbb{Z}^+$, the smallest integer is $n = 12$.
Checking: $S_{11} = 3(2^{11} - 1) = 3(2047) = 6141 < 10000$, and $S_{12} = 3(2^{12} - 1) = 3(4095) = 12285 > 10000$.
Answer: $n = 12$
Sum of Infinite Convergent Geometric Sequences ($S_\infty$)
- Convergence Criterion: An infinite geometric series $u_1 + u_1 r + u_1 r^2 + \dots$ has a finite sum if and only if the terms decay to zero: $$|r| < 1 \iff -1 < r < 1$$ If $|r| \ge 1$, the series diverges and the sum to infinity does not exist.
- Derivation of the Formula: As $n \to \infty$, $r^n \to 0$ because $|r| < 1$. $$S_\infty = \lim_{n \to \infty} \frac{u_1(1 - r^n)}{1 - r} = \frac{u_1(1 - 0)}{1 - r} = \frac{u_1}{1 - r}$$
- Converting Recurring Decimals into Exact Fractions: Any recurring decimal can be decomposed into an infinite geometric series: $$0.\dot{3}\dot{6} = 0.36 + 0.0036 + 0.000036 + \dots = \frac{36}{100} + \frac{36}{10000} + \dots$$ Here $u_1 = \frac{36}{100}$ and $r = \frac{1}{100} \implies S_\infty = \frac{36/100}{1 - 1/100} = \frac{36/100}{99/100} = \frac{36}{99} = \frac{4}{11}$.
Example 4
Evaluating an Infinite Geometric Sum
An infinite geometric series has first term $u_1 = 54$ and fourth term $u_4 = 16$.
(a) Find the common ratio $r$.
(b) Justify why the sum to infinity $S_\infty$ exists.
(c) Calculate the exact value of $S_\infty$.
- Part (a): Find $r$ from $u_4 = u_1 r^3$: $$16 = 54 r^3 \implies r^3 = \frac{16}{54} = \frac{8}{27}$$ Taking the cube root: $$r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}$$
- Part (b): Justify convergence: Since $|r| = \left|\frac{2}{3}\right| = \frac{2}{3} < 1$, the terms decay toward zero and the series converges to a finite sum.
- Part (c): Evaluate $S_\infty = \frac{u_1}{1 - r}$: $$S_\infty = \frac{54}{1 - \frac{2}{3}} = \frac{54}{\frac{1}{3}} = 54 \times 3 = 162$$
Answer: (a) $r = \frac{2}{3}$; (b) Converges since $|r| < 1$; (c) $S_\infty = 162$
Example 5
Logarithmic Threshold Solving & Infinite Sum
A geometric sequence has first term $u_1 = 1200$ and common ratio $r = 0.85$.
(a) Find the smallest value of $n$ such that the $n$th term is strictly less than $50$. [4 marks]
(b) Calculate the exact sum to infinity $S_\infty$ of this sequence. [2 marks]
- Part (a): Set up the inequality $u_n < 50$: $$u_1 r^{n-1} < 50 \implies 1200(0.85)^{n-1} < 50 \quad [M1]$$ $$(0.85)^{n-1} < \frac{50}{1200} = \frac{1}{24} \approx 0.04167 \quad [A1]$$
- Take natural logarithms of both sides: $$\ln\left[(0.85)^{n-1}\right] < \ln\left(\frac{1}{24}\right) \implies (n - 1)\ln(0.85) < \ln\left(\frac{1}{24}\right) \quad [M1]$$ Critical Step: Since $0 < 0.85 < 1$, $\ln(0.85) \approx -0.1625 < 0$. Dividing by a negative flips the inequality: $$n - 1 > \frac{\ln(1/24)}{\ln(0.85)} = \frac{-3.17805}{-0.16252} \approx 19.555$$ $$n > 20.555 \implies n = 21 \quad [A1]$$
- Part (b): State convergence and evaluate $S_\infty$: Since $|r| = |0.85| = 0.85 < 1$, the series converges [R1]. $$S_\infty = \frac{u_1}{1 - r} = \frac{1200}{1 - 0.85} = \frac{1200}{0.15} = \frac{1200}{3/20} = 1200 \times \frac{20}{3} = 8000 \quad [A1]$$
Answers: (a) $n = 21$; (b) $S_\infty = 8000$
Examiner Commentary: Forgetting to flip the inequality sign when dividing by $\ln(0.85)$ is one of the most frequent errors on Paper 2. Always verify by checking: $u_{20} = 1200(0.85)^{19} \approx 54.1 > 50$, whereas $u_{21} = 1200(0.85)^{20} \approx 46.0 < 50$.
A geometric sequence starts $5, 15, 45, 135, \dots$
(a) State the common ratio $r$.
(b) Calculate the $7$th term $u_7$.
(c) Find the sum of the first $6$ terms $S_6$.
View Complete Solution
- Part (a): $r = \frac{15}{5} = 3$.
- Part (b): $$u_7 = u_1 r^6 = 5 \times 3^6 = 5 \times 729 = 3645$$
- Part (c): $$S_6 = \frac{5(3^6 - 1)}{3 - 1} = \frac{5(729 - 1)}{2} = \frac{5(728)}{2} = 5 \times 364 = 1820$$
Final Answer: (a) $r = 3$; (b) $u_7 = 3645$; (c) $S_6 = 1820$
An alternating geometric sequence begins $64, -32, 16, -8, \dots$
(a) Find the common ratio $r$.
(b) Find the $10$th term as an exact fraction.
(c) Calculate the sum to infinity $S_\infty$.
View Complete Solution
- Part (a): $r = \frac{-32}{64} = -\frac{1}{2}$.
- Part (b): $$u_{10} = 64 \left(-\frac{1}{2}\right)^9 = 2^6 \times \left(-\frac{1}{2^9}\right) = -\frac{2^6}{2^9} = -\frac{1}{2^3} = -\frac{1}{8}$$
- Part (c): Since $|r| = \left|-\frac{1}{2}\right| = \frac{1}{2} < 1$: $$S_\infty = \frac{u_1}{1 - r} = \frac{64}{1 - \left(-\frac{1}{2}\right)} = \frac{64}{1 + \frac{1}{2}} = \frac{64}{\frac{3}{2}} = \frac{128}{3}$$
Final Answer: (a) $r = -\frac{1}{2}$; (b) $u_{10} = -\frac{1}{8}$; (c) $S_\infty = \frac{128}{3}$
In a geometric sequence, the $2$nd term is $12$ and the $5$th term is $324$. Find the first term $u_1$ and the common ratio $r$.
View Complete Solution
- Write equations: $$u_2 = u_1 r = 12 \quad \text{--- (1)}$$ $$u_5 = u_1 r^4 = 324 \quad \text{--- (2)}$$
- Divide (2) by (1): $$\frac{u_1 r^4}{u_1 r} = \frac{324}{12} \implies r^3 = 27 \implies r = 3$$
- Substitute into (1): $$u_1 (3) = 12 \implies u_1 = 4$$
Final Answer: $u_1 = 4, r = 3$
Find the exact value of the sum $\sum_{k=1}^{8} 3 \times 2^k$.
View Complete Solution
- Identify the sequence parameters:
First term ($k = 1$): $u_1 = 3 \times 2^1 = 6$.
Second term ($k = 2$): $u_2 = 3 \times 2^2 = 12$.
Common ratio: $r = \frac{12}{6} = 2$.
Number of terms: $n = 8$. - Use finite sum formula: $$S_8 = \frac{6(2^8 - 1)}{2 - 1} = 6(256 - 1) = 6(255) = 1530$$
Final Answer: $1530$
Using the formula for the sum of an infinite geometric series, express the recurring decimal $0.\dot{4}\dot{5} = 0.454545\dots$ as an exact fraction in simplest form.
View Complete Solution
- Decompose into an infinite geometric series: $$0.454545\dots = 0.45 + 0.0045 + 0.000045 + \dots = \frac{45}{100} + \frac{45}{10000} + \dots$$
- Identify $u_1$ and $r$: $$u_1 = \frac{45}{100}, \quad r = \frac{1}{100}$$ Since $|r| = \frac{1}{100} < 1$, the series converges.
- Apply $S_\infty = \frac{u_1}{1 - r}$: $$S_\infty = \frac{\frac{45}{100}}{1 - \frac{1}{100}} = \frac{\frac{45}{100}}{\frac{99}{100}} = \frac{45}{99}$$
- Simplify by dividing numerator and denominator by 9: $$\frac{45 \div 9}{99 \div 9} = \frac{5}{11}$$
Final Answer: $\frac{5}{11}$
The three terms $x - 1$, $x + 2$, and $3x$ are consecutive terms of a geometric sequence with positive terms, in that order. Find the value of $x$.
View Complete Solution
- Apply the geometric mean condition $b^2 = ac$: $$(x + 2)^2 = (x - 1)(3x)$$
- Expand both sides: $$x^2 + 4x + 4 = 3x^2 - 3x$$
- Rearrange into standard quadratic form: $$2x^2 - 7x - 4 = 0$$
- Factor the quadratic: $$(2x + 1)(x - 4) = 0 \implies x = -\frac{1}{2} \quad \text{or} \quad x = 4$$
- Check positive terms condition:
If $x = -\frac{1}{2}$, terms are $-\frac{3}{2}, \frac{3}{2}, -\frac{3}{2}$ (not all positive).
If $x = 4$, terms are $3, 6, 12$ (all positive, ratio $r = 2$).
Therefore, $x = 4$.
Final Answer: $x = 4$
A geometric sequence has first term $u_1 = 500$ and common ratio $r = 0.8$. Find the smallest value of $n$ such that the sum of the first $n$ terms exceeds $2490$.
View Complete Solution
- Note that $S_\infty = \frac{500}{1 - 0.8} = \frac{500}{0.2} = 2500$: Since $2490 < 2500$, such an $n$ exists.
- Set up $S_n > 2490$: $$\frac{500(1 - 0.8^n)}{1 - 0.8} > 2490$$ $$2500(1 - 0.8^n) > 2490$$
- Isolate $0.8^n$: $$1 - 0.8^n > \frac{2490}{2500} = 0.996$$ $$0.8^n < 1 - 0.996 = 0.004$$
- Take natural logarithms (noting $\ln 0.8 < 0$, flip inequality): $$n \ln(0.8) < \ln(0.004) \implies n > \frac{\ln(0.004)}{\ln(0.8)}$$ $$n > \frac{-5.52146}{-0.22314} \approx 24.74$$
- Smallest integer: $n = 25$.
Final Answer: $n = 25$
A ball is dropped from an initial height of $12\text{ metres}$. Each time it bounces, it rebounds to $70\%$ of the height from which it fell.
(a) Find the height the ball reaches on its $5$th rebound.
(b) Find the total vertical distance the ball has travelled when it hits the ground for the $6$th time.
(c) Assuming the ball bounces infinitely, calculate the theoretical total vertical distance it travels before coming to rest.
View Complete Solution
- Part (a): Rebound heights form a GP with $h_1 = 12 \times 0.7 = 8.4\text{ m}, r = 0.7$: $$h_5 = 8.4 \times (0.7)^4 = 12 \times (0.7)^5 \approx 2.01684\text{ m} \approx 2.02\text{ m}$$
- Part (b): Initial drop is $12\text{ m}$. Rebounds $1$ to $5$ travel up and down ($2 \times h_k$): $$\text{Total} = 12 + 2 \times S_5 = 12 + 2 \times \left[ \frac{8.4(1 - 0.7^5)}{1 - 0.7} \right]$$ $$\frac{8.4}{0.3} = 28 \implies S_5 = 28(1 - 0.16807) = 28(0.83193) = 23.294\text{ m}$$ $$\text{Total} = 12 + 2(23.294) = 12 + 46.588 = 58.588\text{ m} \approx 58.6\text{ m}$$
- Part (c): Total infinite distance: $$\text{Distance} = 12 + 2 \times S_\infty = 12 + 2\left(\frac{8.4}{1 - 0.7}\right) = 12 + 2(28) = 12 + 56 = 68\text{ metres}$$
Final Answer: (a) $2.02\text{ m}$; (b) $58.6\text{ m}$; (c) $68\text{ m}$
An infinite geometric series has sum to infinity $S_\infty = 32$ and second term $u_2 = 6$.
(a) Show that $32r^2 - 32r + 6 = 0$.
(b) Find the two possible values of $r$, and the corresponding values of $u_1$.
View Complete Solution
- Part (a): Set up equations: $$S_\infty = \frac{u_1}{1 - r} = 32 \implies u_1 = 32(1 - r)$$ $$u_2 = u_1 r = 6$$ Substitute $u_1$: $$32(1 - r)r = 6 \implies 32r - 32r^2 = 6$$ $$32r^2 - 32r + 6 = 0 \quad \text{[Shown]}$$
- Part (b): Divide by 2 and solve:
$$16r^2 - 16r + 3 = 0$$
$$(4r - 1)(4r - 3) = 0 \implies r = \frac{1}{4} \quad \text{or} \quad r = \frac{3}{4}$$
Both satisfy $|r| < 1$.
If $r = \frac{1}{4}$: $u_1 = 32\left(1 - \frac{1}{4}\right) = 32\left(\frac{3}{4}\right) = 24$.
If $r = \frac{3}{4}$: $u_1 = 32\left(1 - \frac{3}{4}\right) = 32\left(\frac{1}{4}\right) = 8$.
Final Answer: (a) [Shown]; (b) $(r = \frac{1}{4}, u_1 = 24)$ or $(r = \frac{3}{4}, u_1 = 8)$
The first, third, and ninth terms of an arithmetic sequence with non-zero common difference $d$ form the first three consecutive terms of a geometric sequence with common ratio $r \neq 1$.
(a) Show that $u_1 = d$.
(b) Find the common ratio $r$ of the geometric sequence.
(c) Given that the sum of the first $5$ terms of the geometric sequence is $726$, find $u_1$ and $d$.
View Complete Solution
- Part (a): Write the AP terms and equate ratios: $T_1 = u_1, \quad T_2 = u_1 + 2d, \quad T_3 = u_1 + 8d$ Since the terms form a geometric sequence, the common ratio is constant: $\frac{u_1 + 2d}{u_1} = \frac{u_1 + 8d}{u_1 + 2d} \implies (u_1 + 2d)^2 = u_1(u_1 + 8d)$ $u_1^2 + 4u_1 d + 4d^2 = u_1^2 + 8u_1 d$ $4d^2 = 4u_1 d$ Since $d \neq 0$, dividing both sides by $4d$ gives: $u_1 = d \quad \text{[Shown]}$
- Part (b): Find $r$: $$r = \frac{T_2}{T_1} = \frac{u_1 + 2d}{u_1} = \frac{d + 2d}{d} = \frac{3d}{d} = 3$$
- Part (c): Use sum of GP $S_5 = 726$: $$S_5 = \frac{T_1(3^5 - 1)}{3 - 1} = \frac{u_1(243 - 1)}{2} = \frac{242 u_1}{2} = 121 u_1 = 726$$ $$u_1 = \frac{726}{121} = 6$$ Since $u_1 = d$, $d = 6$.
Final Answer: (a) [Shown with 9th term]; (b) $r = 3$; (c) $u_1 = 6, d = 6$
General Term & Ratio
- Common Ratio: $$r = \frac{u_{k+1}}{u_k}$$
- The $n$th Term: $$u_n = u_1 r^{n-1}$$
- Geometric Mean: $$b^2 = ac \iff b = \pm \sqrt{ac}$$
Geometric Series Formulas
- Finite Series Sum: $$S_n = \frac{u_1(1 - r^n)}{1 - r} = \frac{u_1(r^n - 1)}{r - 1} \quad (r \neq 1)$$
- Convergence Criterion: $$|r| < 1 \iff -1 < r < 1$$
- Sum to Infinity: $$S_\infty = \frac{u_1}{1 - r} \quad (|r| < 1)$$
Exam Method Protocols
- Ratio of Arbitrary Terms: $$\frac{u_q}{u_p} = r^{q-p}$$
- Logarithmic Solving for $n$: $$r^n = K \implies n = \frac{\ln K}{\ln r}$$
- Infinite Recurring Decimal: $$0.\dot{a}\dot{b} = \frac{ab}{99}, \quad 0.\dot{a}\dot{b}\dot{c} = \frac{abc}{999}$$
Critical IB Examination Pitfalls in Geometric Sequences
- Applying $S_\infty$ When $|r| \ge 1$: Attempting to calculate a sum to infinity when $r = 1.5$ or $r = -2$ shows fundamental lack of understanding. The series diverges; state clearly that $S_\infty$ does not exist!
- Even Power Sign Ambiguity ($r^2$ or $r^4$): If $r^2 = 9$, $r = \pm 3$. Never discard the negative root without an explicit syllabus or question reason (such as "all terms are positive").
- The Exponent $(n - 1)$ vs $n$: Writing $u_n = u_1 r^n$ is an automatic method penalty. The first term is $u_1 r^0 = u_1$.
- Inequality Flip with Logarithms: When $|r| < 1$, $\ln r$ is negative (e.g. $\ln 0.5 \approx -0.693$). Dividing by $\ln r$ MUST flip the inequality symbol ($\le$ becomes $\ge$).
- Fractional Base Power Formatting: Write $\left(\frac{2}{3}\right)^n$, never $\frac{2^n}{3}$ or $\frac{2}{3}^n$, which risks misinterpretation by examiners.
Question 1
What is the $6$th term of a geometric sequence with $u_1 = 5$ and $r = -2$?
Question 2
For which of the following common ratios does the sum to infinity $S_\infty$ exist?
Question 3
An infinite geometric series has $u_1 = 20$ and $r = 0.6$. What is $S_\infty$?
Question 4
The positive numbers $2, k, 18$ form consecutive terms of a geometric progression. What is the value of $k$?
A geometric sequence has first term $u_1 = 18$ and second term $u_2 = 12$.
View Markscheme
$r = \frac{u_2}{u_1} = \frac{12}{18} = \mathbf{\frac{2}{3}}$ A1
(b)
$S_4 = \frac{18\left(1 - (2/3)^4\right)}{1 - 2/3} = \frac{18(1 - 16/81)}{1/3}$ M1
$= 54\left(\frac{65}{81}\right) = \frac{2 \times 65}{3} = \mathbf{\frac{130}{3}}$ A1
(c)
State convergence condition: $|r| = \frac{2}{3} < 1$ R1
$S_\infty = \frac{u_1}{1 - r} = \frac{18}{1 - 2/3}$ M1
$= \frac{18}{1/3} = \mathbf{54}$ A1
A tree plantation is established with $2500$ trees in year 1. Each year, due to disease, the tree population decreases by $8\%$, so that the number of trees surviving in any year is $92\%$ of the previous year's total.
View Markscheme
$u_1 = 2500, r = 0.92$
$u_6 = 2500(0.92)^5$ M1
$= 2500(0.65908) \approx 1647.7 \implies \mathbf{1648\text{ trees}}$ A1
(b)
Set $u_n < 1000 \implies 2500(0.92)^{n-1} < 1000$ M1
$(0.92)^{n-1} < 0.40$ A1
$(n - 1)\ln(0.92) < \ln(0.40) \implies n - 1 > \frac{\ln(0.40)}{\ln(0.92)}$ M1
$n - 1 > \frac{-0.91629}{-0.08338} \approx 10.989 \implies n > 11.989$
The population first falls below $1000$ in $\mathbf{\text{Year } 12}$ A1
Unit Overview