SOLUTION: Given the arithmetic progression ;3,8,13...Find the least number of terms whose sum is greater than 80..

Question 1080886: Given the arithmetic progression ;3,8,13...Find the least number of terms whose sum is greater than 80..

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
One can just write them out
3+8+13=24
+18=42
+23=65
+28=93, so the 6th term will sum to more than 80.
The formula
S=(n/2)(2a+(n-1)d)
80=(n/2)(6+5(n-1))=(n/2)(6+5n-5)
160=n(5n+1)
5n^2+5n-160=0
n^2+n-32=0
does not factor
n=(1/2)(-1+/- sqrt (1+128)); sqrt 129=11.36
n=10.36/2=5.18. This must be rounded to the next highest number or 6.

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