What is the recursive formula for this sequence?
[tex]\[12, 16, 20, 24, 28, \ldots\][/tex]

A. [tex]\(\left\{\begin{array}{l}a_1=4 \\ a_n=a_{n-1}+12\end{array}\right.\)[/tex]

B. [tex]\(\left\{\begin{array}{l}a_1=12 \\ a_n=a_{n-1}+4\end{array}\right.\)[/tex]

C. [tex]\(\left\{\begin{array}{l}a_1=32 \\ a_n=a_{n-1}+4\end{array}\right.\)[/tex]

D. [tex]\(\left\{\begin{array}{l}a_1=12 \\ a_n=a_{n-1}-4\end{array}\right.\)[/tex]



Answer :

To determine the recursive formula for the given sequence [tex]\(12, 16, 20, 24, 28, \ldots\)[/tex], let's break down the problem step-by-step:

1. Identify the First Term:

The first term of the sequence is given as [tex]\(a_1 = 12\)[/tex].

2. Determine the Common Difference:

To find the common difference ([tex]\(d\)[/tex]), we need to subtract the first term from the second term.
[tex]\[ d = a_2 - a_1 = 16 - 12 = 4 \][/tex]
This indicates that 4 is added to each term to get the next term in the sequence.

3. Formulate the Recursive Formula:

The recursive formula for an arithmetic sequence takes the form:
[tex]\[ a_n = a_{n-1} + d \][/tex]
Given that the first term [tex]\(a_1 = 12\)[/tex] and the common difference [tex]\(d = 4\)[/tex], we can write the recursive formula as follows:
[tex]\[ \left\{\begin{array}{l} a_1 = 12 \\ a_n = a_{n-1} + 4 \end{array}\right. \][/tex]

4. Match with Given Options:

Compare the derived formula with the provided options:
- A. [tex]\( \left\{\begin{array}{l}a_1=4 \\ a_n=a_{n-1}+12\end{array}\right. \)[/tex]: This option has an incorrect first term and common difference.
- B. [tex]\( \left\{\begin{array}{l}a_1=12 \\ a_n=a_{n-1}+4\end{array}\right. \)[/tex]: This matches exactly with our derived formula.
- C. [tex]\( \left\{\begin{array}{l}a_1=32 \\ a_n=a_{n-1}+4\end{array}\right. \)[/tex]: This option has an incorrect first term.
- D. [tex]\( \left\{\begin{array}{l}a_1=12 \\ a_n=a_{n-1}-4\end{array}\right. \)[/tex]: This option has an incorrect common difference.

Hence, the correct recursive formula for the sequence is:
[tex]\[ \text{B.} \left\{\begin{array}{l}a_1=12 \\ a_n=a_{n-1}+4\end{array}\right. \][/tex]