Question

which recursive formula can be used to define the sequence for n>1? 10, 13, 16, 19, 22, 25, ... f(1) = 10, f(n) = f(n - 1)·(7/3); f(1) = 10, f(n) = f(n - 1) + 7; f(1) = 10, f(n) = f(n - 1) + 3; f(1) = 5, f(n) = f(n - 1) + 8

Explanation:

Step1: Analyze the sequence

The given sequence is \(10, 13, 16, 19, 22, 25, \dots\). Let's find the common difference between consecutive terms.
\(13 - 10 = 3\), \(16 - 13 = 3\), \(19 - 16 = 3\), and so on. So the common difference \(d = 3\).

Step2: Check the initial term and recursive relation

The first term \(f(1)\) should be \(10\) (since the first term of the sequence is \(10\)). For a recursive formula of an arithmetic sequence (where we add a common difference), the formula is \(f(n)=f(n - 1)+d\) where \(d\) is the common difference. Here \(d = 3\), so the recursive formula should be \(f(1)=10\) and \(f(n)=f(n - 1)+3\) for \(n>1\).

Let's check the options:

• First option: Multiplies by \(\frac{7}{3}\), not adding, so incorrect.
• Second option: Adds \(7\), but our common difference is \(3\), so incorrect.
• Third option: \(f(1) = 10\) and \(f(n)=f(n - 1)+3\), which matches our analysis.
• Fourth option: Initial term is \(5\) (not \(10\)) and adds \(8\) (not \(3\)), so incorrect.

Answer:

C. \(f(1) = 10, f(n) = f(n - 1) + 3\) (assuming the third option is labeled as C, if the options are labeled A, B, C, D as per the order: A. \(f(1) = 10, f(n) = f(n - 1)\cdot\frac{7}{3}\), B. \(f(1) = 10, f(n) = f(n - 1)+7\), C. \(f(1) = 10, f(n) = f(n - 1)+3\), D. \(f(1) = 5, f(n) = f(n - 1)+8\))