Question

The 20^th term from the end of the progression \[ 20, 19, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots, -129 \frac{1}{4} \] is:

• A. –118
• B. –115
• C. –110
• D. –100

Answer 1

The Correct Option is B

1. To ascertain the 20^th term from the end of the provided arithmetic sequence, we first identify its characteristics. The sequence is: \(20, 19 \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots, -129 \frac{1}{4}\).
2. The differences between the initial terms confirm it is an arithmetic sequence. The second term is \(19 \frac{1}{4}\). This reveals a common difference of \(d = -\frac{3}{4}\) between successive terms.
3. The formula for the \(n^{th}\) term of an arithmetic sequence is \(a_n = a + (n-1) \cdot d\), where \(a\) is the first term (20) and \(d\) is the common difference (\(-\frac{3}{4}\)).
4. To locate the 20^th term from the end, we must first determine the total number of terms (\(n\)) in the sequence. Using the first and last terms: \(-129 \frac{1}{4} = 20 + (n-1) \cdot \left(-\frac{3}{4}\right)\).
5. Solving for \(n\): The last term, \(-129 \frac{1}{4}\), is equivalent to \(-\frac{517}{4}\). The equation becomes \(-\frac{517}{4} = 20 - \frac{3}{4}(n-1)\). Rearranging: \(-\frac{517}{4} - 20 = -\frac{3}{4}(n-1)\), which simplifies to \(-\frac{597}{4} = -\frac{3}{4}(n-1)\). Multiplying both sides by \(-\frac{4}{3}\) yields \(n-1 = 199\), so \(n = 200\).
6. Thus, the sequence contains 200 terms. The 20^th term from the end is equivalent to the \(200 - 20 + 1 = 181^{st}\) term from the beginning. We calculate \(a_{181} = 20 + (181-1) \cdot \left(-\frac{3}{4}\right)\).
7. Calculating \(a_{181}\): \(a_{181} = 20 + 180 \cdot \left(-\frac{3}{4}\right)\). This simplifies to \(a_{181} = 20 - \frac{540}{4}\), then \(a_{181} = 20 - 135\), resulting in \(a_{181} = -115\).
8. The 20^th term from the end of the sequence is therefore -115.