Posted by Bethany on .
The sum of integrs from 4060, inclusive, is 1050. What is the aum of the integers from 6080, inclusive?
I figured out the answer by adding them with a calculator but i was wondering if there was an algebraic way of solving this?

Math 
Steve,
Since you are adding 20 to each number, just add 20*21 to the total.
Or, thinking of the values as an arithmetic progression, recall that the sum of n terms is
Sn = n/2 (T1+Tn)
With a=40,d=1,
S21 = 21/2 (40+60) = 1050
With a=60,d=1,
S21 = 21/2 (60+80) = 1470
Note that 1470 = 1050 + 21*20 
Math 
Bethany,
Ok, thank you!