What is the (n)th term of the sequence (3,6,9,12,\ldots)?

Correct answer: (a_n=3n)

The direct answer is option A: \(a_n=3n\). The sequence is \(3,6,9,12,\ldots\), so each term is a multiple of 3. The first term is \(3=3\times1\), the second is \(6=3\times2\), the third is \(9=3\times3\), and the fourth is \(12=3\times4\). Therefore the term in position \(n\) is \(3\times n\), or \(a_n=3n\). Option A gives exactly this rule. Option B, \(n+3\), gives 4 when \(n=1\), not the first term 3. Option C, \(2n+1\), gives 3 for the first term but gives 5 for the second instead of 6, so it fails. Option D, \(3n-1\), gives 2 for the first term and therefore does not match the sequence. A useful additional check is that the common difference is 3, and an arithmetic sequence beginning at 3 with difference 3 has formula \(a_n=a_1+(n-1)d=3+3(n-1)=3n\). Memory cue: for the positive multiples of 3 beginning with 3, use three times the position number.