Question 367043
First term is a1 = -49000, common difference is d = 1, so an = -49000+(n-1)*1 = -49,000+n-1 = n -49,001.  If an = 50,000, then 50,000 = n - 49,001, or n = 99,001.  So there are 99,001 terms in the sequence.  The sum of all these terms is {{{(99001/2)*(-49000 + 50000) = 99001*500}}}.  The average of all these numbers is then {{{(99001*500)/99001 = 500}}}