find the number n such that the sum of integers from 1 to n-1 is equal to the sum of the integers from n+1 to 49

Question

kropot72 · Answer

Sum of integers from 1 to n-1 = 1/2(n - 1) * (1 + (n - 1)) = 1/2 * n(n - 1)
Sum of integers from n + 1 to 49 = 1/2 * (49 - n) * (n + 50)
1/2 * n(n - 1) = 1/2 * (49 - n)(n + 50) 
$$2n ^{2}-2450=0$$
$$n ^{2}=1225$$
$$n=\sqrt{1225}$$

kropot72 · Answer

The formula used is:
Sum of n terms of an AP = 1/2 * (number of terms) * (sum of first and last terms)

anonymous · Answer

how did you now that n-1 is the number of terms?

kropot72 · Answer

By testing. If n = 4 there are 3 terms. If n = 5 there are 4 terms. If n = 6 there are 5 terms.