SOLUTION: How many terms should we add to exceed 2335 when we add - 18 - 11

SOLUTION: How many terms should we add to exceed 2335 when we add - 18 - 11 - 4 ...?

Question 1099354: How many terms should we add to exceed 2335 when we add - 18 - 11 - 4 ...?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
2335=(n/2)(a1+an)
4670=n(a1+an)
an=a1+d(n-1)=-18+7(n-1)
substitute
4670=n(-18+7n-7-18)=-43n+7n^2
7n^2-43n-4670=0
n=(1/14)(43+/- sqrt (43^2+28(4670)); sqrt term is equal to 364.16
n=(1/14)(407.16)=29.08, round up to 30
an=-18+7(29)=185
2335<=(29/2)(-18+185)=(14.5*167)=2421.5
-18
-11
-4
3
10
17
24
31
38
45
52
59
66
73
80
87
94
101
108
115
122
129
136
143
150
157
164
171
178
185
middle is half way between 15th and 16th
That is 83.5
multiply that by 30=2505
sum of first 29 is 2505-185=2325.
ANSWER IS 30 TERMS

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