Change in arithmetic and geometric sequences — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: constant common difference for arithmetic sequences, constant common ratio for geometric sequences, difference operator behavior, average rate of change over intervals, comparing linear vs exponential change, and recursive/explicit formula applications for AP Precalculus questions.

You should already know: Basic sequence notation and recursive/explicit forms, Average rate of change for continuous functions, Basic exponent and logarithm algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Change in arithmetic and geometric sequences?

This topic, worth approximately 3–5% of the total AP Precalculus exam score per the official College Board CED, falls within Unit 2: Exponential and Logarithmic Functions, and appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections. A sequence is a discrete function whose domain is restricted to a subset of the integers, so analyzing change in sequences follows the same core logic as analyzing change for continuous functions, but adapted to a discrete input set. Arithmetic sequences are defined by constant additive change between consecutive terms, while geometric sequences are defined by constant proportional (multiplicative) change between consecutive terms. This topic acts as a critical bridge: it connects the linear change you studied in Unit 1 to the exponential change that is the core focus of Unit 2. On the exam, you will be asked to identify whether a sequence is arithmetic, geometric, or neither; calculate average and consecutive change over multiple intervals; compare long-term growth rates of arithmetic and geometric sequences; and model real-world discrete processes like periodic savings or population growth with these sequences.

2. Change in Arithmetic Sequences

An arithmetic sequence is a discrete linear sequence defined by the constant additive change between consecutive terms, called the common difference $d$. The sequence difference operator, which gives the change between term $n$ and term $n+1$, is written as: $$\Delta a_n=a_{n+1}-a_n$$ By definition, for any arithmetic sequence, $\Delta a_n=d$ for all $n$ — this constant consecutive change is the defining property of an arithmetic sequence. The explicit form of an arithmetic sequence starting at $n=0$ is $a_n=a_0+nd$, or $a_n=a_1+(n-1)d$ if starting at $n=1$. For any interval from $n=k$ to $n=m$ (where $k<m$), the average rate of change of the sequence is: $$\frac{\Delta a}{\Delta n}=\frac{a_m-a_k}{m-k}=\frac{(a_0+md)-(a_0+kd)}{m-k}=d$$ This means the average rate of change over any interval is also constant, equal to the common difference, just as the slope of a continuous linear function is constant everywhere. This matches the core property of linear change: equal changes in input produce equal changes in output.

Worked Example

Given the sequence defined explicitly by $a_n=7-4n$, find (1) the total change between the 3rd and 8th term, and (2) the average rate of change over the interval $n=1$ to $n=5$, then confirm it matches the common difference.

1. First confirm the sequence is arithmetic: it is linear in $n$, so the common difference is the coefficient of $n$, $d=-4$.
2. Calculate the terms needed: $a_3=7-4(3)=-5$, $a_8=7-4(8)=-25$.
3. Total change = $a_8-a_3=-25-(-5)=-20$. Verify with the change formula: total change = $d(8-3)=-4(5)=-20$, which matches.
4. Calculate average rate of change from $n=1$ to $n=5$: $a_1=7-4(1)=3$, $a_5=7-4(5)=-13$. Average rate of change = $\frac{a_5-a_1}{5-1}=\frac{-13-3}{4}=-4$.
5. The average rate of change matches the common difference $d=-4$, as expected for an arithmetic sequence.

Exam tip: When a problem gives you non-consecutive terms of an arithmetic sequence, calculate the common difference directly as $d=\frac{a_j-a_i}{j-i}$ without solving for the initial term first, to save time on MCQs.

3. Change in Geometric Sequences

Geometric sequences are discrete exponential sequences defined by constant proportional (multiplicative) change between consecutive terms, called the common ratio $r$. By definition, for any geometric sequence: $$\frac{a_{n+1}}{a_n}=r\text{for all }n$$ Unlike arithmetic sequences, the additive change between consecutive terms of a geometric sequence is not constant: $$\Delta a_n=a_{n+1}-a_n=r{a}_n-a_n=a_n(r-1)$$ Additive change is proportional to the current value of the sequence, which is the discrete equivalent of the continuous exponential property that the derivative of an exponential function is proportional to the function's current value. For any two terms separated by $k$ positions, the proportional change is constant: $\frac{a_{n+k}}{a_n}=r^k$, a useful property for solving for $r$ when given non-consecutive terms. The explicit form of a geometric sequence starting at $n=0$ is $a_n=a_0{r}^n$, or $a_n=a_1{r}^{n-1}$ for sequences starting at $n=1$. The average rate of change over an interval depends on the interval, unlike the constant average rate of change for arithmetic sequences.

Worked Example

A geometric sequence has $a_2=18$ and $a_5=486$, with all positive terms. Find the common ratio $r$, then calculate the additive change between $a_3$ and $a_6$.

1. Use the proportional change property for non-consecutive terms: $\frac{a_5}{a_2}=r^{5-2}=r^3$.
2. Substitute the given values: $\frac{486}{18}=27=r^3$, so $r=\sqrt[3]{27}=3$ (we discard the negative root because all terms are positive).
3. Find the required terms: $a_3=a_2\cdot r=18\cdot 3=54$, $a_6=a_5\cdot r=486\cdot 3=1458$.
4. Calculate additive change: $\Delta a=a_6-a_3=1458-54=1404$.
5. Verify with the proportional change rule: $\Delta a=a_3(r^3-1)=54(27-1)=54\cdot 26=1404$, which confirms the result.

Exam tip: If the problem does not specify all terms are positive, remember that $r^k=c$ (for positive $c$ and even $k$) has two solutions: $r=\sqrt[k]{c}$ and $r=-\sqrt[k]{c}$. Do not forget the negative solution unless explicitly told to rule it out.

4. Comparing Long-Term Change in Arithmetic vs Geometric Sequences

A key AP Precalculus skill is comparing the long-term behavior of increasing arithmetic (discrete linear) and increasing geometric (discrete exponential) sequences as $n$ grows large. For any increasing arithmetic sequence ($d>0$, $a_n$ grows linearly) and any increasing geometric sequence ($r>1$, $a_n$ grows exponentially), exponential growth will always outpace linear growth for sufficiently large $n$, even if the arithmetic sequence is larger for small values of $n$. This is a core property that is tested regularly in both MCQs and FRQs, and it often requires using logarithms to solve for the minimum $n$ where the geometric sequence surpasses the arithmetic sequence, connecting this topic to the logarithm content of Unit 2.

Worked Example

Arithmetic sequence $A(n)=100+20n$ and geometric sequence $G(n)=50(1.12{)}^n$ are defined for all integers $n\ge 0$. What is the minimum integer $n$ such that $G(n)>A(n)$?

1. Set up the inequality we need to solve: $50(1.12{)}^n>100+20n$.
2. Test small values to see the trend: at $n=0$, $G=50<100=A$; at $n=15$, $G\approx 50(5.4736)=273.68<100+300=400=A$; at $n=25$, $G\approx 50(17.000)=850>100+500=600=A$.
3. Check $n=24$ to confirm it is the minimum: $G(24)=\frac{G(25)}{1.12}\approx \frac{850}{1.12}\approx 758.9$, $A(24)=100+20(24)=580$, wait — check $n=20$: $G(20)\approx 50(9.6463)=482.3$, $A(20)=100+400=500$, so $G(20)<A(20)$.
4. Check $n=21$: $G(21)=482.3(1.12)\approx 540.2$, $A(21)=100+420=520$, so $G(21)>A(21)$.
5. The minimum integer $n$ is 21, since $n=20$ still has $G(n)<A(n)$.

Exam tip: When asked for the minimum $n$ where a geometric sequence exceeds an arithmetic sequence, always check the integer one below your candidate value. Exams regularly include the candidate value one above the correct answer as a distractor for MCQs.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Treating the common ratio of a geometric sequence as the additive change between consecutive terms, e.g. writing $\Delta a_n=r$ instead of $a_n(r-1)$. Why: Students confuse the defining multiplicative property of geometric sequences with the question's request for "change", which is always additive unless explicitly stated otherwise. Correct move: Underline whether the question asks for "change" (additive) or "common ratio" (multiplicative) before starting to solve.
• Wrong move: Calculating the common difference of an arithmetic sequence as $a_j-a_i$ for non-consecutive terms, e.g. for $a_2=5$, $a_5=14$, calculating $d=14-5=9$. Why: Students are used to finding $d$ from consecutive terms, so they forget to divide by the gap in indices. Correct move: Always divide the difference in term values by the difference in term positions to get $d$, regardless of whether terms are consecutive.
• Wrong move: Stopping at the first crossing of $G(n)>A(n)$ and assuming $G(n)$ stays larger, even when $0<r<1$. Why: Students forget that geometric sequences with $0<r<1$ decay to zero, so they can exceed an increasing arithmetic sequence for small $n$ then fall behind permanently. Correct move: Check the value of $r$ first: if $r<1$ and $A(n)$ is increasing, note that $G(n)$ will eventually fall back below $A(n)$.
• Wrong move: Off-by-one errors when using $a_n=a_1+(n-1)d$ or $a_n=a_1{r}^{n-1}$, e.g. calculating the 5th term as $a_1{r}^5$. Why: Confusion between sequences starting at $n=0$ vs $n=1$ is extremely common. Correct move: Write down explicitly whether the sequence starts at $n=0$ or $n=1$ before plugging into any explicit formula.
• Wrong move: Calculating total change for an arithmetic sequence over $k$ steps as $d$ instead of $kd$. Why: Students confuse consecutive change (change over 1 step) with change over multiple steps. Correct move: Always multiply the common difference by the difference in indices to get total change for arithmetic sequences.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

A geometric sequence has $a_1=10$, $a_3=90$, and $a_5=810$, with all positive terms. What is the average rate of change of the sequence from $n=2$ to $n=4$? A) 60 B) 120 C) 180 D) 240

Worked Solution: First, use the property of geometric sequences that the ratio between terms separated by $k$ positions is $r^k$. We have $\frac{a_3}{a_1}=r^{3-1}=r^2=\frac{90}{10}=9$, so $r=3$ (we discard $r=-3$ because all terms are positive). Next, calculate $a_2=a_{1}r=10\ast 3=30$ and $a_4=a_1{r}^3=10\ast 27=270$. Average rate of change is $\frac{a_4-a_2}{4-2}=\frac{270-30}{2}=120$. The correct answer is B.

Question 2 (Free Response)

Consider two sequences defined for all integers $n\ge 0$: Arithmetic: $A(n)=200+15n$ Geometric: $G(n)=50(1.12{)}^n$ (a) Show that the difference between consecutive terms of $A(n)$ is constant, and find the value of this constant. (b) Calculate $A(n)$ and $G(n)$ for $n=0,10,20,30$, and state whether $G(n)>A(n)$ at each value. (c) Find the smallest integer $n$ such that $G(n)>A(n)$, and justify your answer.

Worked Solution: (a) The difference between consecutive terms is $\Delta A(n)=A(n+1)-A(n)=[200+15(n+1)]-[200+15n]=15$. This is constant for all $n$, so the constant difference is 15. (b) Calculations:

• $n=0$: $A=200$, $G=50$ → $G<A$
• $n=10$: $A=350$, $G\approx 155.3$ → $G<A$
• $n=20$: $A=500$, $G\approx 482.3$ → $G<A$
• $n=30$: $A=650$, $G\approx 1498.0$ → $G>A$ (c) Check $n=21$: $G(21)=482.3\ast 1.12\approx 540.2$, $A(21)=200+15(21)=515$, so $G(21)>A(21)$. Confirm $n=20$: $G(20)\approx 482.3<500=A(20)$. The smallest integer $n$ is 21.

Question 3 (Application / Real-World Style)

A new bakery is tracking daily customer count starting from opening day ($n=0$, opening day). One business analyst models growth as arithmetic: the bakery gains 15 new customers every day, starting from 80 customers on opening day. A second analyst models growth as geometric: daily customer count increases by 4% every day, starting from 150 customers on opening day. After how many full opening days will the geometric model first predict a higher customer count than the arithmetic model? Round to the nearest full day, and interpret the result.

Worked Solution: Write the explicit formulas for each model, where $n$ = number of full opening days after opening day:

• Arithmetic: $A(n)=80+15n$
• Geometric: $G(n)=150(1.04{)}^n$ Test values to find the minimum $n$ where $G(n)>A(n)$:
• $n=36$: $G(36)\approx 150(1.04^{36})\approx 615.5$, $A(36)=80+15(36)=620$, so $G(36)<A(36)$
• $n=37$: $G(37)\approx 615.5\ast 1.04\approx 640.1$, $A(37)=80+15(37)=635$, so $G(37)>A(37)$

After 37 full opening days, the geometric growth model first predicts a higher daily customer count than the constant growth arithmetic model. In context, this means that after ~7 weeks of opening, the percentage-based daily growth model forecasts more customers than the constant daily increase model.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational introduction to discrete exponential change, which is the core of Unit 2: Exponential and Logarithmic Functions. Next, you will extend the properties of discrete geometric sequences to continuous exponential functions, where the multiplicative change property translates to the key property that the rate of change of an exponential function is proportional to the function's value. Without mastering the difference between additive constant change (arithmetic/linear) and multiplicative constant change (geometric/exponential), you will struggle to model continuous exponential growth and decay, solve exponential equations, or compare linear and exponential models in real-world contexts. This topic also lays the groundwork for the study of geometric series later in the unit, which are used for applications like compound interest and annuity calculations.

Properties of exponential functions Exponential vs linear modeling Logarithmic properties and solving exponential equations Finite geometric series