Question

which function rule can be used to find the nth term in the following sequence? 3, 7, 11, 15, 19, ... f(n) = 4n + 19 f(n) = 4n - 1 f(n) = n + 4 f(n) = 2n + 1

Explanation:

Step1: Identify Sequence Type

The sequence 3, 7, 11, 15, 19... is arithmetic (common difference \(d = 4\)). The formula for the \(n\)-th term of an arithmetic sequence is \(f(n)=a_1+(n - 1)d\), where \(a_1 = 3\), \(d = 4\).

Step2: Derive the Formula

Substitute \(a_1 = 3\), \(d = 4\) into the formula:
\(f(n)=3+(n - 1)\times4\)
Simplify: \(f(n)=3 + 4n-4=4n - 1\).

Step3: Verify with Options

• For \(f(n)=4n + 19\): At \(n = 1\), \(4(1)+19 = 23

eq3\).

• For \(f(n)=n + 4\): At \(n = 1\), \(1 + 4 = 5

eq3\).

• For \(f(n)=2n + 1\): At \(n = 1\), \(2(1)+1 = 3\), but at \(n = 2\), \(2(2)+1 = 5

eq7\).

• For \(f(n)=4n - 1\): At \(n = 1\), \(4(1)-1 = 3\); \(n = 2\), \(4(2)-1 = 7\); \(n = 3\), \(4(3)-1 = 11\) (matches the sequence).

Answer:

\(f(n)=4n - 1\) (the option with \(f(n)=4n - 1\))