First of all, it usually does not makes much sense to include the end condition, the idea of a range is to perform hops until the end value is reached.

Nevertheless, you can for instance use itertools.chain for that:

from itertools import chain

for i in chain(range(0,99,20), [99]):
    # ...
    pass

chain concatenates iterables together. So after the range(..) is exhausted, it will iterate over the next iterable, and so on.

The above approach is not very elegant: it requires some knowledge about how chain works. We can however encapsulate that logic:

def range_with_end(start, stop, step):
    return chain(range(start, stop, step), (stop,))

there is a discussion of this here: Iteration over list slices

this is one of the shortest versions:

import numpy as np

lst = np.arange(start = 19, stop = 67, step = 2)
bin_width = 5
search = 50

for ibin in zip(*(iter(lst),) * bin_width):
    print(ibin)
    if min(ibin) <= search <= max(ibin):
        print('found!')
    # or this? not sure what you want...
    if ibin[0] <= search <= ibin[-1]:
        print('found!')

this prints

(19, 21, 23, 25, 27)
(29, 31, 33, 35, 37)
(39, 41, 43, 45, 47)
(49, 51, 53, 55, 57)
found!
found!

Consider tracking the time it takes the code to run (a timer() function), then sleeping for 15 - exec_time seconds after completion.

start = datetime.now()
do_many_important_things()
end = datetime.now()

exec_time = end - start
time.sleep(15-exec_time.total_seconds())