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Thread: word problems sequences and series

  1. #1
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    word problems sequences and series

    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
    __________________________________________________ _________

    6) The total number of digits 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. #2
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    Re: word problems sequences and series

    Quote Originally Posted by Aka
    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...
    I'll give you a hand with this one. It's getting late.(yawn)

    [tex]\text{Use the arithmetic series thing\\

    For the positive:}[/tex]

    [tex]\L\\S_{n}=\frac{n}{2}[2a_{1}+(n-1)d][/tex]

    [tex]a_{1}=2\\d=2[/tex]

    [tex]\L\\h=\frac{n}{2}[2(2)+(n-1)(2)][/tex]

    [tex]\text{odds would be:}[/tex] [tex]\L\\k=\frac{n}{2}[2(1)+(n-1)(2)][/tex]

    [tex]\text{Now subtract h and k. Your answer is there among the choices.}[/tex]

  3. #3
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    Re: word problems sequences and series

    Hello, Aka!

    Here are a few of them . . .

    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?

    [tex]a)\;35\qquad\qquad b)\;34\qquad\qquad c)\:33\qquad\qquad d)\;32\qquad\qquad e)\;31[/tex]
    Your reasoning is correct . . .
    We'll use: [tex]\,t_n\:=\:t_1\,+\,d(n\,-\,1)[/tex]

    We have: [tex]\,t_1\,=\,888,\;\;d\,=\,-7[/tex]

    So: [tex]\,t_n\:=\:888\,-\,7(n\,-\,1)[/tex]

    Solve for [tex]n:\;\;n\:=\:\frac{888\,-\,t_n}{7}\,+\,1[/tex]

    Since [tex]n[/tex] is positive integer, [tex]888\,-\,t_n[/tex] must be divisible by 7.

    And only [tex]t_n\,=\,34[/tex] works . . . answer (b)


    2. The sum of the first [tex]n[/tex] even postive integers is [tex]h[/tex].
    The sum of the first [tex]n[/tex] odd positive integers is [tex]k[/tex].
    Then [tex]h\,-\,k[/tex] is equal to:

    [tex]a)\;\frac{n}{2}\qquad\qquad b)\;\frac{n}{2}\,-\,1\qquad\qquad c)\;n\qquad\qquad d)\;-n\qquad\qquad e)\;n\;-\,1[/tex]
    I'll solve this from square-one . . .


    The sum of the first [tex]n[/tex] even integers: [tex]\,2\,+\,4\,+\,6\,+\,\cdots\,+\,2n[/tex]

    [tex]\;\;[/tex] is an arithmetic series with [tex]t_1\,=\,2[/tex] and [tex]d\,=\,2[/tex]

    Hence: [tex]\,h\;=\;S_{\text{even}}\;=\;\frac{n}{2}[2\cdot2\,+\,2(n-1)] \;=\;n(n\,+\,1)[/tex]


    The sum of the first [tex]n[/tex] odd integers: [tex]\,1\,+\,3\,+\,5\,+\,+\,\cdots\,+\,(2n-1)[/tex]

    [tex]\;\;[/tex]is an arithmetic series with [tex]t_1\,=\,1[/tex] and [tex]d\,=\,2[/tex]

    Hence: [tex]\,k\;=\;S_{\text{odd}}\;=\;\frac{n}{2}[2\cdot1\,+\,2(n-1)] \;= \;n^2[/tex]


    Therefore: [tex]\,h\,-\,k\;=\;n(n\,+\,1)\,-\,n^2\;=\;n[/tex] . . . answer (c)


    3. the largest four-digit number to be foung in the arithmetic sequence
    1,6,11,16,21, ... is >
    Your answer (9996) is correct!

    We have an arithmetic sequence with [tex]t_1\,=\,1,\;d\,=\,5[/tex]

    The [tex]n^{th}[/tex] term is: [tex]\,t_n\;=\;1\,+\,5(n\,-\,1)[/tex]

    We want [tex]t_n[/tex] to be less than 10,000 (a five-digit number).

    So we have: [tex]\,1\,+\,5(n\,-\,1)\;<\;10,000[/tex]

    [tex]\;\;[/tex]Subtract 1: [tex]\,5(n\,-\,1)\;<\;9,999[/tex]

    [tex]\;\;[/tex]Divide by 5: [tex]\,n\,-\,1\;<\;1999.8[/tex]

    [tex]\;\;[/tex]Add 1: [tex]\,n \;<\;2000.8[/tex]

    So, we let [tex]\,n\,=\,2000[/tex]

    Then: [tex]t_{_{2000}}\;=\;1\,+\,5(1999)\;=\;9996[/tex] . . . There!
    I'm the other of the two guys who "do" homework.

  4. #4
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    Re: word problems sequences and series

    Quote Originally Posted by Aka
    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
    Dividing each number of the series by 7 leaves a remainder of 6.

    Which of ) 35 b) 34 c) 33 d) 32 e) 31 leaves a remander of 6 when divided by 7?

    34/7 = 4 + 6 remainder.
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

  5. #5
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    Re: word problems sequences and series

    Quote Originally Posted by Aka
    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...
    The sum of the first n even integers is n(n + 1).

    The sum of the first n odd integers is n^2.

    Therefore, the difference between the first n even integers and the first n odd integers is n(n + 1) - n^2 = n^2 + n - n^2 = n.
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

  6. #6
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    Re: word problems sequences and series

    Quote Originally Posted by Aka
    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?
    9999/5 = 1999.8 or 1999 + 4

    Therefore, the highest 4 digit number in the series is 1999 - 4 = 1996.

    Every number leaves a remainder of 1 when divided by 5.

    The number of terms is 1996/5 = 399.2 or 399 with a 1 remainder
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

  7. #7
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    Re: word problems sequences and series

    Hello, Aka!

    4. The sum of 50 consecutive intergers is 3250. tha largest of these intergers is:

    [tex]a)\;64\qquad\qquad b)\;66\qquad\qquad c)\;112\qquad\qquad d)\;114\qquad\qquad e)\;115[/tex]
    The answer is "None of These"!


    We have an arithmetic series: [tex]a\,+\,(a+1)\,+\,(a+2)\,+\,\cdots\,+\,(a+49)[/tex]

    The sum of the first [tex]n[/tex] terms is: [tex]\,S_n\;=\;\frac{n}{2}[2a_1\,+\,d(n-1)][/tex]

    Our first term is [tex]a[/tex], the common difference is [tex]d = 1,\;n\,=\,50,\;S_{50}\,=\,3250[/tex]

    We have: [tex]\,3250\;=\;\frac{50}{2}[2\cdot a\,+\,1(50\,-\,1)][/tex]

    [tex]\;\;[/tex]then: [tex]\,3250\;=\;25[2a\,+\,49)\;\;\Rightarrow\;\;130\;=\;2a\,+\,49\;\; \Rightarrow\;\;81\;=\;2a[/tex]

    But this gives us: [tex]\,a\;=\;40.5[/tex] . . . and [tex]a[/tex] is supposed to be an integer.

    So I assume there is a typo in the statement of the problem.


    5. January 1, 1986, was a Wednesday.
    January 1, 1992, was what day of the week?
    There is no "formula" for this problem; you must do some Thinking.

    From 01/01/86 to 01/01/92 is six years: [tex]\,6\,\times\,365\:=\:2190[/tex] days.
    But 1988 was a leap year, so there are: 2191 days.

    Since [tex]2191\,\div\,7\;=\;313[/tex] with no remainder,
    [tex]\;\;[/tex]then 01/01/92 is exactly 313 weeks after 01/01/86.

    Therefore, January 1, 1992 is also on a Wednesday.


    6) The total number of digits used to number all the pages of a book was 216.
    Find the number of pages in the book.
    There is no formula for this one either . . . You must baby-talk your way through it.

    Pages 1 to 9: nine 1-digit numbers = 9 digits.

    Pages 10 to 99: ninety 2-digit numbers = 180 digits.

    There are: [tex]\,216\,-\,9\,-\,90\:=\:27[/tex] digits to go.

    These are taken up by the first nine 3-digit numbers: 100, 101, 102, ... , 108

    Therefore, the last page is number 108.
    I'm the other of the two guys who "do" homework.

  8. #8
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    Re: word problems sequences and series

    Quote Originally Posted by Aka
    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
    Assuming the first and last numbers are x and y,
    (x + y)50/2 = 3250 making x +y = 130
    For 50 integers, x - y = 51
    x + y = 130
    x - y = 51
    2x = 181 making x 90.5

    since we are supposed to be dealing with integers, some given information is wrong.
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

  9. #9
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    Thanks for everyones help!

    Sorry, I typed #4 wrong its

    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

  10. #10
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    That makes a world of difference then.

    [tex]\L\\25(2x+98)=3250[/tex]

    [tex]x=16[/tex]

    The first number of the series is 16. Can you figure out what the last one is now?.

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