The sum of 25 consecutive integers is 1000. What is the smallest integer used as an addendum?
let the first term be a
clearly d = 1
sum(n) = (n/2)(2a + (n-1)d)
1000 = (25/2)(2a + 24(1))
2000 = 25(2a + 24)
2000= 50a + 600
50a = 1400
a = 1400/50 = 28
check:
sum(25) = (25/2)( 56 + 24)
= 12.5(80) = 1000
To find the smallest integer used as an addendum, we need to first understand the given information.
The sum of 25 consecutive integers is 1000. This means we have a series of 25 numbers that are in consecutive order, and when we add them all together, the sum is 1000.
Let's denote the smallest integer in the series as "x". Since we have 25 consecutive integers, the next consecutive integer will be (x + 1), the one after that will be (x + 2), and so on. The 25th integer will be (x + 24).
To find the sum of the 25 consecutive integers, we can use the formula for the sum of an arithmetic series:
Sum = (Number of Terms / 2) * (First Term + Last Term)
In this case, the number of terms is 25, the first term is x, and the last term is (x + 24). So we can write the equation:
1000 = (25 / 2) * (x + (x + 24))
Simplifying the equation:
1000 = (25 / 2) * (2x + 24)
Now we can solve for x:
1000 = 12.5 * (2x + 24)
80 = 2x + 24
2x = 80 - 24
2x = 56
x = 56 / 2
x = 28
Therefore, the smallest integer used as an addendum is 28.