Question

The 20th 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

Answer: (B)

–115

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