9514 1404 393
Answer:
A. d = ∑[k=1,n] (4.5+0.5k)
B. 9
Step-by-step explanation:
A) The series has first term 5 and common difference 0.5. The general term is given by ...
an = a1 +d(n -1)
an = 5 + 0.5(n -1) . . . . substitute numbers for this series
an = 4.5 +0.5n . . . . . . simplify
The sum is the sum of these terms:
[tex]\displaystyle \boxed{d_n=\sum_{k=1}^n(4.5+0.5k)}[/tex]
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B) The sum of terms of an arithmetic series is the average of the first and last, multiplied by the number of terms. We want to solve for n to make the sum greater than 60.
[tex]\displaystyle d_n=\frac{n}{2}(5 +4.5 +0.5n)\\\\\dfrac{n}{2}(9.5+0.5n)>60\\\\n(19+n)>240\qquad\text{multiply by 4}\\\\n^2+19n+9.5^2>240+9.5^2\qquad\text{complete the square}\\\\n+9.5>\sqrt{330.25}\qquad\text{square root}\\\\n>18.17-9.5=8.67\qquad\text{positive root, subtract 9.5}[/tex]
Melissa's total distance will first exceed 60 miles the 9th time she trains.
