Which sequence is an AP with common difference (\frac{1}{4})?

Correct answer: (\frac{1}{2},\frac{3}{4},1,\frac{5}{4})

An arithmetic progression must have the same difference between every two consecutive terms. The required difference here is \(\frac14\), so subtract each term from the next and check whether the result is always \(\frac14\). Option D follows the pattern of adding one-fourth at every step, so it is the correct sequence.

For option D, \(\frac34-\frac12=\frac14\), \(1-\frac34=\frac14\), and \(\frac54-1=\frac14\). All consecutive differences are equal. In option A, the differences are not all equal; option B also has changing differences, and option C doubles irregularly. Therefore only option D is an AP with common difference \(\frac14\).