Question

Sum of integers between 10-50

Answer 1

41 integers, so
41/2 (2*10 + 40) = _____

Answer 2

Case1 : you are including 10 and 50
there are 41 terms
sum = n(first + last)/2 = 41(10+50)/2 = 1230
or
consider it an arithmetic series
a = 10, d = 1 , term(n) = 50 = a+(n-1)d
50 = 10+ n-1
n = 41, as before
sum = (41/2)(20 + 40(1)) = 1230

case2 : use normal definition of "between" and exclude the 10 and 50
so add all the numbers from 11 to 49, so there are 39 of those
sum = 39/2 (22 + 38(1)) = 1170
or
a = 11, d=1, term(n) = 49
49 = 11 + n-1
n = 39
sum = 39/2 (22 + 38(1)) = 1170

or
we could have just subtracted 10 and 50 from 1230 to get 1170

You decide which interpretation to use, I would go with case2