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 determine the 20^th term from the end of the given arithmetic sequence, let's first identify the characteristics of the sequence. The sequence is given as \(20, 19 \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots, -129 \frac{1}{4}\). 
2. Check the first few differences to confirm it's an arithmetic sequence:
   • Second term: \(19 \frac{1}{4}\)
   • This indicates a common difference, \(d = -\frac{3}{4}\), between terms (i.e., the difference between subsequent terms is negative).
3. The formula for the \(n^{th}\) term of an arithmetic sequence is given by: \(a_n = a + (n-1) \cdot d\), where:
   • \(a\) is the first term, \(20\) in this case.
   • \(d\) is the common difference, \(-\frac{3}{4}\).
4. To find the 20^th term from the end, first determine the total number of terms in the sequence. Use the first term and last given term: \(-129 \frac{1}{4} = 20 + (n-1) \cdot \left(-\frac{3}{4}\right)\).
5. Solve for \(n\):
   • Rewriting \(-129 \frac{1}{4}\) as \(-\frac{517}{4}\)
   • \(-\frac{517}{4} = 20 - \frac{3}{4}(n-1)\)
   • \(-\frac{517}{4} - 20 = -\frac{3}{4}(n-1)\)
   • \(-\frac{597}{4} = -\frac{3}{4}(n-1)\)
   • Multiply by \(-4/3\): \(n-1 = 199\)
   • \(n = 200\)
6. So, there are 200 terms in the sequence. To find the 20^th term from the end, calculate the 181^st term from the start: \(a_{181} = 20 + (181-1) \cdot \left(-\frac{3}{4}\right)\)
7. Calculate \(a_{181}\):
   • \(a_{181} = 20 - 180 \cdot \frac{3}{4}\)
   • \(a_{181} = 20 - \frac{540}{4}\)
   • \(a_{181} = 20 - 135\)
   • \(a_{181} = -115\)
8. Therefore, the 20^th term from the end of the sequence is -115, which is the correct answer.

Answer 2

To find the 20th term from the end of the given arithmetic progression (A.P.):

Given sequence:
\[ 20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots, -129 \frac{1}{4} \]

The common difference (\(d\)) is calculated as:  
\(d = -1 + \frac{1}{4} = -\frac{3}{4}.\)

To find the 20th term from the end, we consider the reversed A.P. starting from:  
\(a = -129 \frac{1}{4} \quad \text{and} \quad d = \frac{3}{4}.\)

The formula for the \(n\)-th term of an A.P. is given by:  
\(a_n = a + (n - 1)d.\)

Substituting the given values:  
\(a_{20} = -129 \frac{1}{4} + (20 - 1) \cdot \frac{3}{4}.\)

Simplifying:  
\(a_{20} = -129 \frac{1}{4} + 19 \cdot \frac{3}{4}.\)

Combining the terms:  
\(a_{20} = -\frac{517}{4} + \frac{57}{4}.\)
\(a_{20} = -\frac{460}{4}.\) 
\(a_{20} = -115.\)

Conclusion:
The 20th term from the end of the progression is:  \(-115.\)

The Correct Answer is : -115