Question

7, 11, 15, 19,...
what is the recursive formula that you used to create the sequence?
(a) $a_n = a_{n - 1} + 7$; $a_1 = 4$
(b) $a_n = a_{n + 1} + 7$; $a_1 = 4$
(c) $a_n = a_{n - 1} + 4$; $a_1 = 7$
(d) $a_n = a_{n + 1} + 4$; $a_1 = 7$

Explanation:

Step1: Analyze the sequence

The sequence is 7, 11, 15, 19,... Let's check the difference between consecutive terms. 11 - 7 = 4, 15 - 11 = 4, 19 - 15 = 4. So the common difference is 4.

Step2: Check the recursive formula form

A recursive formula for an arithmetic sequence is usually of the form \(a_n=a_{n - 1}+d\), where \(d\) is the common difference and \(a_1\) is the first term. Here, \(d = 4\) and \(a_1=7\). Let's check the options:

• Option A: \(a_n=a_{n - 1}+7,a_1 = 4\) → Incorrect, first term is wrong and difference is wrong.
• Option B: \(a_n=a_{n+1}+7,a_1 = 4\) → Incorrect, form and first term wrong.
• Option C: \(a_n=a_{n - 1}+4,a_1 = 7\) → Matches our analysis (common difference 4, first term 7, and recursive relation \(a_n=a_{n - 1}+4\)).
• Option D: \(a_n=a_{n+1}+4,a_1 = 7\) → Incorrect form (should be \(a_{n - 1}\) not \(a_{n+1}\) for the recursive relation of an increasing arithmetic sequence).

Answer:

C. \(a_n = a_{(n - 1)} + 4; a_1 = 7\)