Finding the 1000th Term in a SequenceDate: 1/19/96 at 15:34:21 From: Anonymous Subject: Number Our class was given the following problem in class: Two kids are on a car trip, and there's nothing to do, so they decide to count telephone poles. One kid counts normally, 1,2,3,4,5...25,26,27...31,32,33, etc. But the other kid counts them a different way. He counts them like this: 1,2,3,4,5,6,7,8,9,10,9,8,7,6,5,4,3,2,1, and then up to twenty, and down again, then thirty, and down, etc. When the normal kid gets to one thousand, their father tells them to,"SHUT UP!" What number was the weird kid on when their father told them to shut up, if both kids were counting at the same time? Date: 1/21/96 at 20:54:44 From: Doctor Ken Subject: Re: Number Hey, neat question! Essentially, you're looking for the 1000th term in the sequence 1,2,3,4,5,6,7,8,9,10,9,8,7,6,5,4,3,2,1,2,...29,30,29,28,...., which looks like it could be pretty hairy. Let's see if we can find a pattern or a formula: 1,2,3,4,5,6,7,8,9,10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 2,...29,30,29,28,...., 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,... See what I did? I lined up the two kids' counting methods next to each other. Notice that when the weird kid starts over at 1, the normal kid hits 19. That's because the weird kid has 10+9=19 terms in the first part of his sequence (until he starts over at 1). The next part of his sequence will be 20+19=39 terms, for a total of 19+39 = 58 terms. If you wanted to, you could just keep going like this, adding 19 + 39 + 59 + 79 + ... until you get just below 1000, and then just do it by hand from there. Or, if you want to be REALLY slick, you could figure out a formula for the sum of those series: a formula where you could plug in the number of groups (a group is a count-up and a count-down), and you'd get the number of terms in the weird kid's sequence until then. For instance, you would plug in 1 and you'd get 19, and you'd plug in 2 and you'd get 19 + 39 = 58, and so on. Here are some hints if you wanted to do it that way: The sequence is also known as 20 + 40 + 60 + ... + 20*n - n. Unless I messed up. 1+2+3+4+5+6+...+n = n(n+1)/2. See if this sheds any light on the problem for you. Good luck! -Doctor Ken, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/