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

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