"Find all n satisfying 1 + 3 + 5 + 7 + ... + 125 = n + (n + 2) + (n + 4) + (n + 6) + ... + 209."
L = 1+3+5+7+...+125 = 3969
R = n+(n+2)+(n+4)+(n+6)+...+209
Write the last term of R as n+2N
R = n*(N+1) + N*(N+1)
Now the last term n + 2N = 209, so N = (209 - n)/2, hence:
R = (n + 209)(211 - n)/4
Equate L and R:
(n + 209)(211 - n)/4 = 3969
I'll leave you to solve this quadratic to find the two values of n.