SOLUTION: If you add the consecutive counting numbers starting with 1, what number will cause the sum to exceed 1000? Explain.

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Question 351050: If you add the consecutive counting numbers starting with 1, what number will cause the sum to exceed 1000? Explain.
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
n=number of terms. a[1]=1st term, a[n]=final term, S=sum of finite arithmetic sequence.
n/2(a[1]+a[n])=S[n]
n/2(1+n)>=1000
n(n+1)>=2000
n^2+n-2000>=0 Quadratic formula (below)
n>=44.22
n=45 the number that will cause the sum to exceed 1,000.
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Ed
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-2000+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A-2000=8001.

Discriminant d=8001 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+8001+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+8001+%29%29%2F2%5C1+=+44.2241545476267
x%5B2%5D+=+%28-%281%29-sqrt%28+8001+%29%29%2F2%5C1+=+-45.2241545476267

Quadratic expression 1x%5E2%2B1x%2B-2000 can be factored:
1x%5E2%2B1x%2B-2000+=+1%28x-44.2241545476267%29%2A%28x--45.2241545476267%29
Again, the answer is: 44.2241545476267, -45.2241545476267. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-2000+%29