Which is the (n)th term of the sequence (2,5,8,11,\ldots)?

Correct answer: (a_n=3n-1)

The direct answer is option A: \(a_n=3n-1\). The sequence 2, 5, 8, 11 increases by 3 each time, so it is an arithmetic progression with first term \(a_1=2\) and common difference \(d=3\). The nth-term formula for an arithmetic progression is \(a_n=a_1+(n-1)d\). Substituting gives \(a_n=2+(n-1)3=2+3n-3=3n-1\). Checking is useful: for \(n=1\), the formula gives 2; for \(n=2\), it gives 5; for \(n=3\), it gives 8. Option A is therefore correct. Option B, \(3n+2\), gives 5 when \(n=1\), so it starts with the wrong first term. Option C, \(2n+3\), gives 5, 7, 9 for the first three values and does not have the required difference of 3. Option D, \(5n-3\), gives 2 initially but then gives 7, not 5, so its common difference is 5. The common mistake is to identify the difference correctly but forget the \(n-1\) in the formula. Memory cue: first term plus one less than the position multiplied by the common difference.