Question

Melissa has started training for a race. The first time she trains, she bikes 5 miles. Each subsequent time she trains, she bikes 0.5 mile farther than she did the previous time.
A. Use summation notation to write an arithmetic series to represent the total distance Melissa has biked after she has trained n times.

B. Use the formula for a partial sum of the series you wrote in part a, above, to answer this question: What is the least number of times Melissa must bike for her total distance biked during training to exceed 60 miles? (3 points)

Answer 1

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:

  $\displaystyle \boxed{d_n=\sum_{k=1}^n(4.5+0.5k)}$

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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.

  $\displaystyle d_n=\frac{n}{2}(5 +4.5 +0.5n)\\\\\dfrac{n}{2}(9.5+0.5n)>60\\\\n(19+n)>240\qquad\text{multiply by 4}\\\\n^2+19n+9.5^2>240+9.5^2\qquad\text{complete the square}\\\\n+9.5>\sqrt{330.25}\qquad\text{square root}\\\\n>18.17-9.5=8.67\qquad\text{positive root, subtract 9.5}$

Melissa's total distance will first exceed 60 miles the 9th time she trains.

Answer 2

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