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,))

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())

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!