Question

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

Answer 1

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)
\[2n ^{2}-2450=0\]
\[n ^{2}=1225\]
\[n=\sqrt{1225}\]

Answer 2

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

Answer 3

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

Answer 4

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