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

$\displaystyle s_{n-1}= \frac{1}{2}(n(n-1)) $

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

I try this: $\displaystyle (n + 1) + (n+2) + (n+3) + ....... + 49 $

a = n+1 common difference = 1

nth term = 2n

thus

49th term $\displaystyle = 2(49) = 98 $

let n = 1

first term = 2(1) = 2

thus sum of n+1 to 49

$\displaystyle s_{49}= \frac{49}{2}(2(2)+(49-1)(1)) = 49(26)=1274 $

thus

$\displaystyle \frac{1}{2}(n(n-1)) = 1274 $

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