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Counting backwards; Sequences

  1. Jun 7, 2006 #1

    Aya

    User Avatar

    Please help I have ALOT of questions!!

    1.)Starting at 888 and counting backward by 7, a student counts 888, 881 and 874, and so on. Which of the following numbers will be included?

    a) 35 b) 34 c) 33 d) 32 e) 31

    Ok, so I by using the calcuator the aswer is 34, but how would you calcuate this using an equation or something more sensible...

    I think I have to use this one

    tn=a + (n-1) d

    a= 888
    n=
    d= -7
    tn=

    ... now what

    _____________________________________________________________

    2. the sum of the first n even postive integers is h. the sum of the first n odd positive integers is k. then h -k is equal to:

    a) n/2
    b) n/2 -1
    c) n
    d) -n
    e) n - 1

    so... nK and nh

    I don't get it...

    ____________________________________________________________

    3. the largest four- digit number to be foung in the arithmetic sequence
    1,6,11,16,21, ... is

    t1= 1
    t2= 6
    t3= 11
    t4= 21


    9996?
    because it goes form 1 to 6 the (last number)
    2nd last number stayes for 2 ed 1,1,2,2
    everything else up one

    but is there an equation for this or something?
    ___________________________________________________________
    4. the sum of 50 consecutive intergers is 3250. tha largest of these intergers is

    a)64
    b)66
    c)112
    d)114
    e)115

    I don't know how to so this...is it the same as before?
    ___________________________________________________________
    5 Jan 1, 1986 was a wed. Jan 1 1992 was waht day of the week?

    86 wed
    87 thurs
    88 fri
    89 sat
    90 sun
    91 mon
    92 tues

    tuesday?

    what would an equatin be
    ___________________________________________________________

    the total number of digets used to number all the pages of a book was 216. Find the number of pages in the book.

    tn= a + (n-1) d
    216= 1 + (n- 1) 1
    216= 1 + n - 1
    216=n

    is that wrong...

    so...
    total number of digits? sum?

    Sn= n/2 [ 2a + ( n-1) d ]
    216= n/2 [ 2(1) + (1-1) 1]
    216= n/2 (2)
    216= n

    umm....?

    Pleas help!
    thanks
     
  2. jcsd
  3. Jun 8, 2006 #2
    1. As you wrote,
    tn = a + (n-1)d
    tn = 888 + (n-1)(-7)
    tn = 895 - 7n
    Check whichever value satisfies.

    2. The odd numbers are 1,3,5,..... What will be the sum of such an A.P.?
    The even numbers are 2,4,6,.... What will be sum of this A.P.?

    3. You are right. Logic is better here than an equation. Another way to think about this is that each number is 1 more than a multiple of 5. So 9995+1 = 9996

    4. Something seems wrong. Is the question typed right?

    5. You are wrong.
    86 wed
    87 thurs
    88 fri
    89 SUN ( because 88 was a leap year. An extra day)
    90 mon
    91 tues
    92 wed
    You may use this method to predict any day of any year
    http://quasar.as.utexas.edu/BillInfo/doomsday.html

    6. You need logic here, not equation.
    From 1-9 ---> 9 digits
    From 10 - 99 (90 numbers) -----> 90 * 2= 180 digits
    So we have counted 189 digits and remaining are 27 digits that need to be counted, i.e. nine numbers.
     
  4. Jun 8, 2006 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Then you got lost in the calculation because the answer is NOT 34. Since you are "counting by 7" the difference between 888 and each number is a multiple of 7: 888- 881= 7, 888- 874= 14= 2*7, etc.
    888-35= 843= 7*120+ 3, not a multiple of 7. 888- 34= 844= 7*120+ 4, 888- 33= 885= 7*120+ 5, 888- 32= 886= 7*120+ 6, 888- 31= 847= 7(121). Only 31 is in the list.
    Did you look at simple examples? The sum of the first three even integers is 2+ 4+ 6= 12. The sum of the first 3 odd integers is 1+ 3+ 5= 9. The difference is 3. \ The sum of the first 4 even integers is 2+ 4+ 6+ 8= 20. The sum of the first 4 odd integers is 1+ 3+ 5+ 7= 16. The difference is 4. The sum of the first 5 even integers is 2+ 4+ 6+ 8+ 10= 30. The sum of the first 5 odd integers is 1+ 3+ 5+ 7+ 9= 25. The difference is 5. That's pretty obvious isn't it?

    As you have given it, that's NOT an arithmetic sequence: Here 6-1= 5, 11- 6= 5, but 21- 11= 10, not 5. You must mean 1, 6, 11, 16, 21, ... This is really like problem 1: you are "counting by 5". Each number in the sequence is 1 more than a multiple of 5. What is the largest 4 digit number that is a multiple of 5? 9995. What is one more than that? 9996.
    This is impossible. The sum of two consecutive numbers is the sum of an odd number and an even number and so must be odd. The sum of 50 consecutive numbers is the sum of 25 such pairs and so must be the sum of 25 odd numbers which is odd. But 3250 is even.
    Are you just making up numbers? That formula, Sn= n/2[2a+ (n-1)d] is correct- and is the first time you've mentioned such a formula for arithmetic sequences. It applies to all the problems you've given! But why would a and n and d all equal 1? a is the starting number of an arithmetic sequence. Since you start numbering pages with 1, yes, a= 1. Yes, the "common difference", since this is just regular counting, is 1: d= 1. But n is what you want to find: (n/2)[2+ (n-1)]= 216
    Solve that for n.
     
  5. Jun 8, 2006 #4

    Aya

    User Avatar

    1.
    are you sure? the answer in the back of the book sayes it is 34 and i cheked it again.
    _____________________________________________________________
    2. ok, thanks
    _______________________________________________________________
    4. the sum of 50 consecutive even intergers is 3250. tha largest of these intergers is

    a)64
    b)66
    c)112
    d)114
    e)115
    _____________________________________________________________
    5. Thanks, I did not know 89 was a leap year
    ____________________________________________________________
    6. (n/2)[2+ (n-1)]= 216
    216/2
    =108_____________________________________________________________
     
  6. Jun 11, 2006 #5

    Curious3141

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    Homework Helper

    34 is the correct answer for the first part.

    888 - 34 = 854 =(7)(122)

    Halls, your subtraction was off by ten. :smile:
     
  7. Jun 11, 2006 #6

    HallsofIvy

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    I really need to review my arithmetic!
     
  8. Jun 12, 2006 #7

    VietDao29

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    Homework Helper

    Uhmm, I just wonder how you can get 108 by solving that equation... :rolleyes:
    The problem asks for the sum of the number of digits, not the sum of the number of each page.
    There are 9 1-digit numbers (from 1 to 9)
    There are 90 2-digit numbers (from 10 to 99)
    There are 900 3-digit numbers (from 100 to 999)
    and so on...
    To number all pages from 1 to 9, we have to use 9 digits, right? And to number all pages from 10 to 99, we have to use 90 x 2 = 180 digits.
    Now, add those 2 numbers together, we have: 180 + 9 = 189 (digits). That means we use 189 digits to number the pages from 1 to 99.
    Since the problem tells that 216 digits are used. We are left with:
    216 - 189 = 27 (digits)
    And all of the following pages must be numbered with 3-digit numbers right? So how many pages that are numbered with a 3-digit number? How many pages does the book have?
    Can you go from here? :)
     
  9. Jun 12, 2006 #8

    BobG

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    Science Advisor
    Homework Helper

    That makes a big difference.

    If you pair the first and last even digit, your job will only be half as big. 1st even digit is 2, 50th even digit is 100; 2nd even digit is 4, 49th even digit is 98; etc.

    You should be able to come up with an equation to express that, plus figure out the sum of the first 50 even digits. (It will obviously be smaller than 3250).

    There's two ways to 'slide' the sequence over.

    1) One easy way is to figure out how the rate of change for each 'slide' to the right. If you slide right one even number, each number in your sequence increases by two.

    2) If you knew the starting and ending point of the sequence, the easiest way to get the sum would be to calculate the sum from 2 to the ending point, calculate the sum from 2 to the last even number before the starting point. Subtract the second sum from the first sum. You could probably work with this to come up with what you want, but the first seems so much easier.
     
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