Posted by **Bethany** on Wednesday, July 2, 2014 at 11:43pm.

The sum of integrs from 40-60, inclusive, is 1050. What is the aum of the integers from 60-80, 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**, Thursday, July 3, 2014 at 12:15am
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**, Thursday, July 3, 2014 at 12:26am
Ok, thank you!

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