Question:

The sum of the integers from 1 to n-1 is equal to the sum of integers from the n + 1 to 49. Find n

I am able to find 1 to n - 1 which is

but for (n+1) + (n+2) + ....... + 49

I try this:

a = n+1 common difference = 1

nth term = 2n

thus

49th term

let n = 1

first term = 2(1) = 2

thus sum of n+1 to 49

thus

thus n = -49.9801....... or n = 50.980...

but n is an integer

please can some point me in the right direction

how do I found the first term, common difference and number of terms for the series from (n+1) to 49

Thank you in advance