If the sum of the integers from 15 to 50, inclusive, is equal to the sum of the integers from n to 5, inclusive, and n < 15, then n =
a. -50
b. -49
c. -35
d. -15
e. -14
Is this question even possible to answer? If you keep on adding consecutive negative integers, won't it be impossible for the final sum to be 1170 (the sum of the integers from 15 to 50)?
Assuming $n < 0$ we get $$ S_n = \sum_{k=n}^5 k = \sum_{k=1}^5 k - \sum_{k=1}^{-n} k = 15 - \frac{-n(-n+1)}{2} < 15 $$ This has to be equated with $$ S = \sum_{k=15}^{50} k = \sum_{k=1}^{50} k - \sum_{k=1}^{14} k = \frac{50\,51}{2} - \frac{14\,15}{2} = 1275 - 105 = 1170 $$ which is not possible for any such $n$.
Assuming a printing error for the upper bound, we might try $$ S_n = \sum_{k=n}^{50} k = \sum_{k=1}^{50} k - \sum_{k=1}^{-n} k $$ this would lead to the equation $$ S = S_n \iff \\ \sum_{k=1}^{50} k - \sum_{k=1}^{14} k = \sum_{k=1}^{50} k - \sum_{k=1}^{-n} k \iff \\ \sum_{k=1}^{-n} k = \sum_{k=1}^{14} k \iff \\ -n(-n+1) = 14 \cdot 15 \\ $$ which gives $n = -14$.
Sum of 15 to 50 is positive. Sum of 0 to 14 is positive but as each term is less than each term of 15 to 50 and there are fewer of them. So the sum of any non-negative number n to 5 or 5 to n with 0 <= n < 15 is less than the sum of 15 to 50.
Adding negative numbers makes things less.
So $\sum_{n < 15} n... 5 \le \sum_{0 \le n < 15} n ... 5 < \sum 15 ... 50$.
No need to actually think.
But if we did:
$\sum_{i=m}^n i = \frac {(m + n)(|n - m| + 1)}{2}$
And thus $n \ge 5 \implies$ $\frac {(5 + n)(n - 4)}{2} = 65*26/2 \implies n^2 - 19n - 1710 = 0$ so $n = \frac {19 + \sqrt{19^2 + 4*1710}}{2} > 15$ or
$\frac {(5 + n)(6 - n)}{2} = 65*26/2 \implies n^2 - n + 1660 = 0$ so $n = \frac {1 \pm \sqrt{1 - 4*1660}}{2} \in \mathbb R$.
