The Type aₙ₊₁ = aₙ + d

This is the simplest type of recurrence relation, and it corresponds directly to an arithmetic sequence.

The Recurrence Relation

Meaning: each term is a fixed constant $d$ more than the previous term.

Solution Method

Starting from the recurrence relation:

$$\begin{aligned} a_2 &= a_1 + d \\ a_3 &= a_2 + d = a_1 + 2d \\ a_4 &= a_3 + d = a_1 + 3d \\ &\vdots \\ a_n &= a_1 + (n-1)d \end{aligned}$$

This is exactly the general term formula of an arithmetic sequence!

Worked Examples

Example 1: Basic Solution

The sequence $\{a_n\}$ satisfies $a_1 = 3$ and $a_{n+1} = a_n + 5$. Find $a_n$.

Solution:

This is of the type $a_{n+1} = a_n + d$, where $d = 5$.

$a_n = a_1 + (n-1)d = 3 + (n-1) \times 5 = 5n - 2$

Example 2: A Practical Application

A company made a profit of 1 million yuan in its first year, and its profit increases by 0.2 million yuan every year thereafter. Find the profit in the $n$th year.

Solution:

Let the profit in the $n$th year be $a_n$ (in ten-thousand yuan). Then:

• $a_1 = 100$
• $a_{n+1} = a_n + 20$

This is an arithmetic-type recurrence with $d = 20$:

$a_n = 100 + (n-1) \times 20 = 20n + 80 \text{ (ten-thousand yuan)}$

Practice Problems

Exercise 1

The sequence $\{a_n\}$ satisfies $a_1 = -2$ and $a_{n+1} = a_n + 3$. Find $a_{10}$.

Exercise 2

The sequence $\{a_n\}$ satisfies $a_1 = 5$ and $a_{n+1} - a_n = -2$. Find the general term formula.

Exercise 3

The sequence $\{a_n\}$ satisfies $a_3 = 10$ and $a_{n+1} = a_n + 4$. Find $a_1$ and the general term formula.

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Summary

Symbols Used in This Article

中英对照