# Change in arithmetic and geometric sequences

> AP Precalculus · Unit 2: Exponential and Logarithmic Functions
> Source: https://www.owlsprep.com/study/ap-precalculus-u2-change-in-arithmetic-and-geometric/

This sub-topic covers constant change properties of arithmetic and geometric sequences, difference operator behavior, average rate of change over discrete intervals, and long-term growth comparisons for AP Precalculus exam questions.

**Prerequisites:** Basic sequence notation and recursive/explicit sequence forms; Average rate of change for continuous functions; Fundamental exponent and logarithm algebra

## Learning objectives

- Identify arithmetic and geometric sequences by their change properties
- Calculate total and average rate of change over discrete intervals
- Solve for common difference/ratio given non-consecutive terms
- Compare long-term growth behavior of arithmetic and geometric sequences

## Core Concepts of Discrete Sequence Change

Sequences are discrete functions with domains restricted to subsets of integers, so analyzing change follows the same logic as continuous functions but adapted to discrete inputs. This topic makes up 3–5% of the total AP Precalculus exam score, appearing in both multiple-choice and free-response sections. It acts as a critical bridge between linear change (Unit 1) and the exponential change that forms the core of Unit 2.

**Arithmetic Sequence** — A discrete sequence defined by constant additive change between consecutive terms. The constant change is called the common difference $d$.

*Example:* $1, 4, 7, 10,...$ with common difference $d=3$

**Geometric Sequence** — A discrete sequence defined by constant proportional (multiplicative) change between consecutive terms. The constant ratio is called the common ratio $r$.

*Example:* $2, 6, 18, 54,...$ with common ratio $r=3$

**Exam command terms**

Common command term interpretations for this topic:

- **Identify sequence type** — You must check consecutive differences (for arithmetic) and consecutive ratios (for geometric) to confirm classification *(If both are non-constant, the sequence is neither arithmetic nor geometric)*

## Change in Arithmetic Sequences

The sequence difference operator gives the additive change between consecutive terms, defined as:

$$\Delta a_n = a_{n+1} - a_n$$

By definition, for any arithmetic sequence, $\Delta a_n = d$ (constant) for all $n$. This constant consecutive change is the defining property of arithmetic sequences. The explicit form of an arithmetic sequence starting at $n=0$ is $a_n = a_0 + nd$, and average rate of change over any interval from $n=k$ to $n=m$ is always constant:

$$\frac{a_m - a_k}{m - k} = d$$

**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 is:
5. $$\frac{a_5 - a_1}{5-1} = \frac{-13 - 3}{4} = -4$$
6. 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.

## Change in Geometric Sequences

Geometric sequences are discrete exponential sequences defined by a constant common ratio $r$, the constant proportional change between consecutive terms. By definition, the ratio of any two consecutive terms is constant:

$$\frac{a_{n+1}}{a_n} = r \quad \text{for all } n$$

Unlike arithmetic sequences, the additive change between consecutive terms of a geometric sequence is not constant:

$$\Delta a_n = a_{n+1} - a_n = a_n(r-1)$$

Additive change is proportional to the current term value, which is the discrete equivalent of the way continuous exponential growth changes in proportion to its value. For non-consecutive terms separated by $k$ positions, the proportional change is constant:

$$\frac{a_{n+k}}{a_n} = r^k$$

The explicit form for a sequence starting at $n=0$ is $a_n = a_0 r^n$, or $a_n = a_1 r^{n-1}$ for sequences starting at $n=1$. Unlike arithmetic sequences, the average rate of change over an interval of a geometric sequence is not constant.

**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.

## Comparing Long-Term Growth

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 ($d>0$) and any increasing geometric sequence ($r>1$), exponential growth will always outpace linear growth for sufficiently large $n$, even if the arithmetic sequence is larger for small values of $n$. This question often requires testing integer values (or using logarithms) to find the minimum $n$ where the geometric sequence surpasses the arithmetic sequence.

**Worked example:** Arithmetic sequence $A(n) = 100 + 20n$ and geometric sequence $G(n) = 50(1.12)^n$ are defined for all integers $n \geq 0$. 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 identify the general trend: at $n=0$, $G=50 < 100 = A$; at $n=15$, $G \approx 273.68 < 400 = A$; at $n=25$, $G \approx 850 > 600 = A$.
3. Narrow down by checking smaller values: at $n=20$, $G(20) \approx 482.3$, $A(20) = 100 + 20(20) = 500$, so $G(20) < A(20)$.
4. Check $n=21$: $G(21) = 482.3(1.12) \approx 540.2$, $A(21) = 100 + 20(21) = 520$, 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 MCQ distractor.

## Common pitfalls

- **Wrong:** 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 it fails: 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: Underline whether the question asks for 'change' (additive) or 'common ratio' (multiplicative) before starting to solve
- **Wrong:** 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 $d=9$
  - Why it fails: Students are used to finding $d$ from consecutive terms, so they forget to divide by the gap in indices
  - Correct: Always divide the difference in term values by the difference in term positions to get $d$, regardless of whether terms are consecutive
- **Wrong:** Stopping at the first crossing of $G(n) > A(n)$ and assuming $G(n)$ stays larger, even when $0<r<1$
  - Why it fails: 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: 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:** 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 it fails: Confusion between sequences starting at $n=0$ vs $n=1$ is extremely common
  - Correct: Write down explicitly whether the sequence starts at $n=0$ or $n=1$ before plugging into any explicit formula
- **Wrong:** Calculating total change for an arithmetic sequence over $k$ steps as $d$ instead of $kd$
  - Why it fails: Students confuse consecutive change (change over 1 step) with change over multiple steps
  - Correct: Always multiply the common difference by the difference in indices to get total change for arithmetic sequences

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Arithmetic consecutive change | $\Delta a_n = a_{n+1} - a_n = d$ | Constant for all $n$, defining property of arithmetic sequences |
| Total change (arithmetic, non-consecutive) | $a_j - a_i = d(j-i)$ | Works for any $i < j$ |
| Average rate of change (arithmetic) | $\frac{a_j - a_i}{j-i} = d$ | Always constant, equal to common difference |
| Geometric common ratio | $\frac{a_{n+1}}{a_n} = r$ | Constant for all $n$, defining property of geometric sequences |
| Consecutive additive change (geometric) | $\Delta a_n = a_n(r-1)$ | Proportional to current term, not constant |
| Ratio of non-consecutive terms (geometric) | $\frac{a_j}{a_i} = r^{j-i}$ | Works for any $i < j$ |
| Explicit form ($n=0$ start) | Arithmetic: $a_n = a_0 + dn$; Geometric: $a_n = a_0 r^n$ | Adjust exponent/offset to $(n-1)$ for $n=1$ start |
| Long-term growth comparison | For $d>0, r>1, a_0>0$: $r^n$ outgrows $dn$ as $n \to \infty$ | Exponential growth always outpaces linear growth for large $n$ |

## What's next

This topic is the foundational introduction to discrete exponential change, the core of AP Precalculus Unit 2: Exponential and Logarithmic Functions. Mastering the difference between additive constant change (arithmetic/linear) and multiplicative constant change (geometric/exponential) is critical for all subsequent topics in this unit. You will next extend these discrete properties to continuous exponential functions, where the multiplicative change property translates to the key rule that the rate of change of an exponential function is proportional to the function's value. This topic also lays the groundwork for geometric series, used for applications like compound interest and annuity calculations, and for comparing linear and exponential models in real-world contexts.

- [Change in linear and exponential functions](https://www.owlsprep.com/study/ap-precalculus-u2-change-in-linear-and-exponential/)
- [Exponential Functions for AP Precalculus](https://www.owlsprep.com/study/ap-precalculus-u2-exponential-functions/)
- [Exponential function manipulation](https://www.owlsprep.com/study/ap-precalculus-u2-exponential-function-manipulation/)

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