Geometric sequences

Read rough values off each graph and test them.

• •(A) about 4, 16, 25, 27, 28, 28. The ratios are roughly 3.7, 1.5, 1.1, 1.0, so they are not constant.
• •(B) falls and then rises. With a positive $r$, a geometric sequence only ever rises or only ever falls.
• •(C) 28, 25, 22, 19, 16, 13. A constant difference of −3, so it is arithmetic.
• •(D) about 27, 14, 7, 4, 2, then close to zero. Each value is about half the one before, so $r \approx 0.5$. This is geometric decay.

Shape alone gets you there: (D) is the only graph whose dots curve and flatten towards zero. The 2020 subject report published no multiple choice results.

Divide the second term by the first:

$$r = \frac{-900}{1000} = -0.9$$

$$\begin{aligned} t_4 &= 1000 \times (-0.9)^{3} \\ &= -729 \end{aligned}$$

Or keep multiplying by −0.9: 1000, −900, 810, −729. A negative ratio flips the sign every term, so here the odd-numbered terms are positive and the even-numbered terms are negative.

(C) −700 continues the sizes 1000, 900, 800, 700 by taking away 100 each time, an arithmetic pattern with alternating signs, and 31.4% of students chose it. (A) has lost the sign. 54.3% chose (D).

Growing at 12% a year gives $r = 1.12$. Ten years of growth is ten multiplications by 1.12, starting from the current population:

$$\begin{aligned} &15\,480 \times 1.12^{10} \\ &= 48\,078.5\ldots \\ &\approx 48\,079 \end{aligned}$$

If you call the current population $t_1$, then after 10 years you want $t_{11} = 15\,480 \times 1.12^{(11-1)}$, the same calculation. Part 4 is about exactly this.

(B) adds 12% of 15 480 ten times, which is simple (arithmetic) growth. (C) is $15\,480 \times 1.2$. (D) is one year of growth. 51.4% of students chose (A).

“Doubles every year” is multiplication by 2, so the rule is geometric with $r = 2$. That rules out (C) and (D), which add 2 fish a year.

January 2022 is $t_1$. Count the doublings to January 2030: one each for 2023, 2024, 2025, 2026, 2027, 2028, 2029 and 2030. That is 8 doublings, so January 2030 is term 9.

$$\begin{aligned} t_9 &= 40 \times 2^{8} \\ &= 10\,240 \approx 10\,000 \end{aligned}$$

With $n = 8$ the rule gives $40 \times 2^{7} = 5120$, only half the capacity. That is the check that catches the slip: substitute and see whether the answer matches the story.

41.4% of students chose (A), more than chose the key. 36.2% chose (B).

Check the ratios:

$$\frac{40\,000}{50\,000} = \frac{32\,000}{40\,000} = 0.8$$

The boat keeps 80% of its value each year. The table starts at 0 years, so four years after purchase is four multiplications by 0.8:

$$50\,000 \times 0.8^{4} = 20\,480$$

In rule form, $t_1 = 50\,000$ at 0 years makes 4 years $t_5$, and $t_5 = 50\,000 \times 0.8^{(5-1)}$ is the same calculation. Or extend the table: 25 600 at 3 years, 20 480 at 4 years.

(D) $25 600 is one term short, $50\,000 \times 0.8^{3}$, and 18.9% chose it. (C) $22 000, chosen by 26.9%, continues the drops of 10 000 and 8000 as 6000 and 4000, shrinking differences instead of a constant ratio. 41.4% chose (B).

Rebounds to 75% gives $r = 0.75$. The 25.6 m is the drop height, the value before any bounce. Each bounce is one multiplication by 0.75, so after the third bounce:

$$25.6 \times 0.75^{3} = 10.8\ \text{m}$$

Listing is just as quick: 19.2 m, 14.4 m, 10.8 m. With the formula-book rule, let $t_n$ be the height after the $n$th bounce, so $t_1 = 19.2$ and $t_3 = 19.2 \times 0.75^{2} = 10.8$.

(C) 14.4 m treats the drop as the first bounce, $25.6 \times 0.75^{2}$, and 25.7% chose it. (D) is the first bounce and (A) is the fourth. 56.3% chose (B).

Here the rule is given, so its $n$ wins: $n = 1$ is the first bounce and $t_1 = 240$.

$$\begin{aligned} t_1 &= 240 \\ t_3 &= 240 \times 0.5^{2} = 60 \\ t_1 - t_3 &= 240 - 60 \\ &= 180\ \text{cm} \end{aligned}$$

Compare 2025 Question 6, where the 25.6 m was the drop height. Here 240 is the first rebound, because the rule says $t_1 = 240$.

(A) 90 is $120 - 30$, which uses the power $n$ instead of $n - 1$. (B) is $t_1 - t_2$ and (D) is $t_1 - t_4$. 65.9% chose (C).

a) Decreasing by 17% leaves 83% each year:

$$r = 1 - 0.17 = 0.83$$

1 mark · correctly determines the common ratio (the guide notes it may be implied by subsequent working)

Define the variables, then write the model. Let $n$ = the number of years since 2015, so $n = 1$ is the beginning of 2016, and let $t_n$ = the number of birds.

$$t_n = 483 \times 0.83^{(n-1)}$$

1 mark · determines geometric model

One mark is for $r = 0.83$, one for the complete model $t_n = 483 \times 0.83^{(n-1)}$. The recursive form $t_n = 0.83 \times t_{n-1}$, $t_1 = 483$ was also accepted.

b) The beginning of 2021 is five years after the beginning of 2016. Five multiplications by 0.83 make it the sixth term, $n = 6$.

1 mark · correctly determines the $n$ value

$$\begin{aligned} t_6 &= 483 \times 0.83^{5} \\ &= 190.255\ldots \end{aligned}$$

1 mark · determines $t_6$

Birds come in whole numbers, so round: about 190 birds are predicted to remain at the beginning of 2021.

1 mark · states a reasonable answer

The guide allowed follow-through for $t_6$ and the final answer. Using $n = 5$ gives 229 birds and costs the $n$ mark.

Step 1 · Find $t_1$ and $r$. Check the ratio for both pairs.

$$\begin{aligned} r &= \frac{71\,060}{83\,600} = \frac{60\,401}{71\,060} \\ &= 0.85 \\ t_1 &= 83\,600 \end{aligned}$$

1 mark · correctly determines the values for $t_1$ and $r$

Step 2 · The model. The question defines $n$ as the number of years since the end of 2020, so $n = 1$ is the end of 2021 and $t_n$ is the value.

$$t_n = 83\,600 \times 0.85^{(n-1)}$$

1 mark · shows use of appropriate method for geometric sequence

Step 3 · The end of 2028 is 8 years after the end of 2020, so $n = 8$.

$$\begin{aligned} t_8 &= 83\,600 \times 0.85^{(8-1)} \\ &= 26\,800.24 \end{aligned}$$

1 mark · predicts value at the end of 2028

Step 4 · One third of the value at the end of 2021.

$$\frac{83\,600}{3} = 27\,866.67$$

1 mark · determines value for the one third comparison

Step 5 · Compare and conclude.

$$26\,800.24 < 27\,866.67$$

Yes. The value at the end of 2028, $26 800.24, will be lower than one third of the value at the end of 2021, $27 866.67.

1 mark · provides decision

The guide says the decision mark “can only be awarded if prior working supports decision”, so a bare “yes” scores nothing. It accepted listing the terms year by year, the recursive form $t_n = r\,t_{n-1}$, and comparing $26\,800.24 \times 3 = 80\,400.72$ with 83 600 instead.

The trap is $n$. Take the end of 2028 as $n = 7$ and you get $31 529.70, above one third, and the decision flips.

Before anything else, ask what one step is. The table goes up in 5-day jumps, so $r$ is a ratio per five days, not per day. The plays are running totals, so the value at day 60 is the total the songwriter is paid for.

Step 1 · Define the variables. Let $n$ = the number of 5-day periods, with $n = 1$ at day 5, and $t_n$ = total plays (thousands) at the end of period $n$.

1 mark · correctly defines the variables

Step 2 · The ratio, checked on every pair.

$$r = \frac{12}{8} = \frac{18}{12} = \frac{27}{18} = 1.5$$

1 mark · correctly determines the parameter $r$

Step 3 · The model, with $t_1 = 8$.

$$t_n = 8 \times 1.5^{(n-1)}$$

1 mark · correctly determines a geometric (exponential) model

Step 4 · Day 60 is $60 \div 5 = 12$ periods, so $n = 12$.

$$\begin{aligned} t_{12} &= 8 \times 1.5^{11} \\ &= 691.98\ \text{thousand plays} \end{aligned}$$

1 mark · determines total number of plays

Step 5 · Income. Watch the units: the rate is in cents, the prediction is in dollars.

$$\begin{aligned} &0.175 \times 691\,980 \\ &= 121\,096.5\ \text{cents} \\ &= \text{\textdollar}1210.97 \end{aligned}$$

1 mark · determines income

Step 6 · Compare and conclude. $1210.97 is more than $1000, so the prediction that the songwriter will be owed at least $1000 is reasonable, if the plays keep growing geometrically. They are predicted to be owed about $211 more than that.

1 mark · evaluates reasonableness of solution

One of the six marks was for defining the variables. The guide accepted other definitions of $n$, equivalent recursion models, the total in plays or in thousands of plays, and money rounded to the nearest dollar, the nearest 5 cents or truncated. The 2021 subject report showed an excerpt with “a different but acceptable evaluation of the prediction”, so the verdict can be worded more than one way, as long as it rests on your numbers.

The three traps in this question:

• •The step size. Treating $r = 1.5$ as a daily ratio gives an answer larger by a factor of about 56 million ($1.5^{44}$). Always ask what one increment of $n$ represents before using the rule. Counting day 60 as $n = 11$ is the smaller version of the same slip: 461.32 thousand plays, $807.31, and the verdict flips.
• •Cents against dollars. The rate is quoted in cents, the prediction in dollars.
• •The answer shape. Calculate, then compare explicitly against the claim, then give the verdict in context. A bare $1210.97 does not score the final mark.

Read the graph. Model 1 starts at $250 in 2020 and reaches $2750 in 2038. Model 2 starts at 0 in 2020 and rises $126 a year.

Define the variables. Let $n$ = the number of years since 2019, so 2020 is $n = 1$, and let $t_n$ = the amount of money. Then 2038 is $n = 19$ and 2030 is $n = 11$.

Model 1 compounds annually, so it is geometric. Substitute the two known terms and solve for $r$:

$$\begin{aligned} t_n &= t_1 r^{(n-1)} \\ 2750 &= 250 \times r^{18} \\ r^{18} &= 11 \\ r &= 1.1425 \end{aligned}$$

1 mark · correctly substitutes the values into a geometric rule

$$t_n = 250 \times 1.1425^{(n-1)}$$

1 mark · determines geometric model for Model 1

Model 2 is arithmetic, with $t_1 = 0$ and $d = 126$:

$$\begin{aligned} t_n &= 0 + (n-1) \times 126 \\ &= 126(n-1) \end{aligned}$$

1 mark · correctly determines an arithmetic model for Model 2

In 2030, $n = 11$:

$$\begin{aligned} \text{Model 1: } t_{11} &= 250 \times 1.1425^{10} \\ &= 947.33 \\ \text{Model 2: } t_{11} &= 126 \times 10 \\ &= 1260 \end{aligned}$$

1 mark · determines the amounts for both models in 2030

$$1260 - 947.33 = 312.67$$

In 2030, Model 2 is $313 more than Model 1.

1 mark · determines difference to nearest dollar

1 mark · shows logical organisation communicating key steps

Keeping $r$ unrounded gives $947.31 and the same $313. The guide accepted other definitions of $n$, such as years since 2000, with models to match, and allowed follow-through. Its logical-organisation mark lists “defining variables, connecting statements, use of equality signs”, so write the working as a chain like the one above.

“$n$ = number of years after 2020” makes 2021 $n = 1$, so the 2021 costs are the first terms. Accommodation costs twice what meals cost.

$$\begin{aligned} m_1 &= c = \text{\textdollar}60 \\ a_1 &= 2c = \text{\textdollar}120 \end{aligned}$$

1 mark · correctly determines values for $m_1$ and $a_1$

2025 is 5 years after 2020, so $n = 5$.

1 mark · correctly determines $n = 5$

$$m_5 = 60 + 3(5-1) = \text{\textdollar}72$$

1 mark · uses arithmetic model to determine daily cost for a person in 2025 for meals ($m_5$)

$$\begin{aligned} a_5 &= 120 \times 1.1^{(5-1)} \\ &= \text{\textdollar}175.69 \end{aligned}$$

1 mark · uses geometric model to determine daily cost for a person in 2025 for accommodation ($a_5$)

$$\begin{aligned} &72 \times 7 + 175.69 \times 7 \\ &= \text{\textdollar}1733.83 \end{aligned}$$

1 mark · calculates total cost for a person in 2025 for 7 days

$1733.83 is between $1500 and $2000, so the estimate is reasonable.

1 mark · provides appropriate statement of reasonableness linked to prior working

The guide accepted 1733.844 or 1734, and listing the terms to the 5th term instead of using the rules. The 2024 subject report praised a statement of reasonableness that compares the total “to the cost estimate values given in the question”, so name both ends of the range.

Using $n = 4$ for 2025 gives $1601.04. That is still inside the range, so the verdict hides the slip, but the $n$ mark is gone.

Test both sequences both ways. The slopes −0.8, 0.4, −0.2 have differences 1.2 and −0.6, which are not constant. Their ratios are:

$$r = \frac{0.4}{-0.8} = \frac{-0.2}{0.4} = -0.5$$

So the slopes are geometric with $t_1 = -0.8$ and $r = -0.5$:

$$t_n = -0.8 \times (-0.5)^{(n-1)}$$

1 mark · correctly determines the geometric sequence parameters for the slopes

The intercepts 1.2, 2.7, 4.2 go up by 1.5 each time, so they are arithmetic with $t_1 = 1.2$ and $d = 1.5$:

$$t_n = 1.2 + (n-1) \times 1.5$$

1 mark · correctly determines the arithmetic sequence parameters for the $y$-intercepts

Line 5 uses term 5 of each sequence:

$$\begin{aligned} m &= -0.8 \times (-0.5)^{4} \\ &= -0.05 \\ c &= 1.2 + 4 \times 1.5 \\ &= 7.2 \end{aligned}$$

1 mark · determines slope for Line 5

1 mark · determines $y$-intercept for Line 5

So Line 5 is $y = -0.05x + 7.2$. Solve it simultaneously with Line 1, $-0.8x + 1.2 = -0.05x + 7.2$, by taking 1.2 from both sides first:

$$\begin{aligned} -0.8x &= -0.05x + 6 \\ -0.75x &= 6 \\ x &= -8 \end{aligned}$$

1 mark · determines $x$-coordinate of intersection point

$$y = -0.8(-8) + 1.2 = 7.6$$

1 mark · determines $y$-coordinate of intersection point

The intersection point is $(-8, 7.6)$.

1 mark · shows logical organisation communicating key steps

Listing the terms to the fifth was accepted instead of the rules: slopes −0.8, 0.4, −0.2, 0.1, −0.05 and intercepts 1.2, 2.7, 4.2, 5.7, 7.2. The negative ratio makes the slope change sign from line to line while shrinking towards zero.