SOLUTION: How many terms of the arithmetic series 1491 + 1484+1477+... are needed to give a sum of zero

Question 656238: How many terms of the arithmetic series 1491 + 1484+1477+... are needed to give a sum of zero
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
In this case, a1 = 1491 and d = -7 (since 1484-1491 = -7)

We want Sn = 0, so let's all plug this into the formula below

Sn = (n*(2*a1 + d*(n-1)))/2

to get

0 = (n*(2*1491 + (-7)*(n-1)))/2

and now let's solve for n.

0 = (n*(2*1491 + (-7)*(n-1)))/2

0 = (n*(2982 -7n + 7))/2

0 = (n*(-7n + 2989))/2

0 = (-7n^2 + 2989n)/2

0 = -7n^2 + 2989n

-7n^2 + 2989n = 0

-7n(n - 427) = 0

-7n = 0 or n - 427 = 0

n = 0 or n = 427

Ignore the solution n = 0 since you can't have 0 terms.

So you need 427 terms.
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