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?
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
Ok, thank you!
Yes, there is indeed an algebraic way to solve this problem. To find the sum of a series of consecutive integers, you can use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is given by: Sn = (n/2)(a + l), where Sn is the sum of the series, n is the number of terms in the series, a is the first term, and l is the last term.
In this case, we want to find the sum of the integers from 60-80, inclusive. To find the first term (a) and the last term (l), we can subtract 39 from 40 and add it to 60. This gives us a = 60 and l = 80.
Next, we can determine the number of terms in the series. Since we want to find the sum of the integers from 60-80, inclusive, there are 21 terms, starting from 60 and ending at 80.
Using the formula, we can find the sum (Sn) as follows:
Sn = (n/2)(a + l)
= (21/2)(60 + 80)
= (21/2)(140)
= 21 * 70
= 1470
Therefore, the sum of the integers from 60-80, inclusive, is 1470.