find the \( n \) the term of the Series \( 2,-5,8,-11,14, \ldots \)

We begin by examining the series:   2, -5, 8, -11, 14, … Step 1. Notice that if we ignore the signs, the magnitudes form:   2, 5, 8, 11, 14, … which is an arithmetic sequence with first term 2 and common difference 3. Thus, the nth term’s absolute value is:   2 + (n - 1) · 3 = 3n - 1 Step 2. Next, observe the pattern of signs:   n = 1 → positive,   n = 2 → negative,   n = 3 → positive, etc. This alternation can be represented using (–1) raised to a power. A convenient choice is:   (–1)^(n+1) Because when n = 1, (–1)^(2) = 1 (positive), and when n = 2, (–1)^(3) = –1 (negative), which fits the series. Step 3. Putting it all together, the general (nth) term is:   aₙ = (–1)^(n+1) (3n – 1) Thus, the nth term of the series is:   aₙ = (–1)^(n+1)(3n – 1)