This story problem describes an arithmetic sequence where n is the day in January and a_{n} is how many miles Liana runs that day. The first term is how many miles Liana runs on the first day: a_{1} = 0.25 The difference is the number of miles by which she increases her run each day: d = 0.5 Since this is an arithmetic sequence, the n^{th} term is a_{n} = a_{1} + (n – 1)d. This is the distance Liana runs on the n^{th} day in January. 1. On January 8th, the 8th day of January, Liana runs a_{8} miles. We use the formula to find a_{8}. a_{8} = a_{1} + (7)d = 0.25 + (7)(0.5) = 3.75 On the 8th day of January, Liana runs 3.75 miles. 2. We want to know the first day on which Liana runs 5 miles. This means we want to know the smallest value of n for which a_{n} ≥ 5. Using the formula for an arithmetic sequence, we want to know for which n we first have a_{1} + (n – 1)d ≥ 5 We put in the values of a_{1} and d and solve the inequality: 0.25 + (n – 1)(.5) ≥ 5 (n – 1)(.5) ≥ 4.75 n – 1 ≥ 9.5 n ≥ 10.5 A sequence doesn't have a 10.5th term, so that can't be the right answer. Let's find a_{10} and a_{11} and see which is a better choice. a_{10} = 0.25 + (9)(.5) = 4.75 a_{11} = 0.25 + (10)(.5) = 5.25 On January 10th Liana runs 4.75 miles, so she didn't run 5 miles that day. On January 11th she ran more than 5 miles. January 11th is the first day she runs 5 miles. The best advice we can give for story problems is to write out the first few terms of the sequence so you can see the pattern. Like avoiding brain freeze from eating ice cream too fast, the only way to get better is to practice. |