2 years ago

A shopkeeper orders a special brand of coffee powder to be delivered to

Mathematics

1 answer:

Minchanka [31]2 years ago

3 0

Answer:

$3800

Step-by-step explanation:

30 x 2 = 60

60 + 3 = one container

60 containers = 63 x 60

3780 + 20

3800

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If the sum of 4040 consecutive integer numbers is 2020, find out the absolute difference between the first number and the last n

bearhunter [10]

Let $x$ be the smallest of the consecutive integers summing to 2020. Then the largest integer in the sum is $x+4039,$ so our answer is $\boxed{4039}.$

For completeness, though, let's solve for the value of $x$ that satisfies the conditions of the problem. Recall that the sum of an arithmetic series with first term $a_1,$ last term $a_n,$ and $n$ terms, is $\frac{n(a_1+a_n)}{2}.$ Plugging in our known values gives us the equation $\frac{4040(2x+4039)}{2}=2020,$ which simplifies to $2x+4039=1.$ Thus, $x=-2019,$ so the 4040 consecutive integers are $-2019,-2018,\dots,-1,0,1,\dots,2018,2019,2020.$ Notice how everything except the $2020$ cancels out!