Find the term of the AP: . Write the recursive rule for this AP.

Understand the Question An Arithmetic Progression (AP) is a sequence where each term after the first is obtained by adding a constant value called the common difference () to the preceding term. For an AP with first term and common difference : The term is given by the explicit formula: . The recursive rule expresses each term in terms of the previous term: , and for . Step 1 · Find the Term Given AP: First term . Common difference : Using the formula for the term, : Step 2 · Write the Recursive Rule A recursive rule defines each term using the previous term: First term: General relation: Thus, the recursive rule is: Answer > ; Recursive rule: Common Mistakes Sign Error in Common Difference: Calculating as instead of . Since the terms are decreasing, must be negative. Incomplete Recursive Rule: Writing only while forgetting to state the initial condition and domain .

Predicting What Comes Next: Sequences and Progressions | Exercise 8.2

Question 3

Find the $n^{\text{th}}$ term of the AP: $11, 8, 5, 2, \dots$ . Write the recursive rule for this AP.

Solution

Understand the Question

An Arithmetic Progression (AP) is a sequence where each term after the first is obtained by adding a constant value called the common difference ( $d$ ) to the preceding term.
For an AP with first term $a$ $a$ and common difference $d$ $d$ :
- The $n^{\text{th}}$ term is given by the explicit formula: $t_n = a + (n - 1)d$ .
- The recursive rule expresses each term in terms of the previous term: $t_1 = a$ , and $t_n = t_{n-1} + d$ for $n \ge 2$ .

Step 1 · Find the $n^{\text{th}}$ Term

Given AP: $11, 8, 5, 2, \dots$ Diagram 1

First term $a = 11$ .

Common difference $d$ :

\begin{aligned} d &= 8 - 11 \\[0.6em] &= -3 \end{aligned}

Using the formula for the $n^{\text{th}}$ term, $t_n = a + (n-1)d$ :

\begin{aligned} t_n &= 11 + (n-1)(-3) \\[0.6em] &= 11 - 3n + 3 \\[0.6em] &= 14 - 3n \end{aligned}

Step 2 · Write the Recursive Rule

A recursive rule defines each term using the previous term:

First term: $t_1 = 11$
General relation: $t_n = t_{n-1} + d$

\begin{aligned} t_n &= t_{n-1} + d \\[0.6em] &= t_{n-1} - 3 \end{aligned}

Thus, the recursive rule is:

t_1 = 11, \quad t_n = t_{n-1} - 3 \quad \text{for } n \ge 2

Answer

$n^{\text{th}} \text{ term: } t_n = 14 - 3n$ ; Recursive rule: $t_1 = 11, \; t_n = t_{n-1} - 3 \text{ for } n \ge 2$

Common Mistakes

Sign Error in Common Difference: Calculating $d$ as $11 - 8 = 3$ instead of $\text{second term} - \text{first term} = 8 - 11 = -3$ . Since the terms are decreasing, $d$ must be negative.
Incomplete Recursive Rule: Writing only $t_n = t_{n-1} - 3$ while forgetting to state the initial condition $t_1 = 11$ and domain $n \ge 2$ .

Question 3

More questions in Exercise 8.2