Tuesday, March 17, 2009

Arithmetic progression questions

A Stadium is built with 200 seats in the first row. After that, each row has 50 more seats than the previous row. How many seats are in the 25th row?

The 15th value in a set that increases by 6 is 250, what is the first value?

What is the sum of the values of a set that contains 23 numbers, begins at 5 and increases by 4?

A housing estate is built with five homes on the first row, and there are 3 more homes on each row. How many homes are there on the 15th row?

If there are 50 boxes on the bottom of a stack which is 15 rows tall, and as the stack grows in height, there are 2 less boxes in each row, how many rows of boxes are there in the first row?


Please post other questions for the class if you do have any to solve.
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10 comments:

  1. sinMarch 18, 2009 at 2:02 PM

    First term is a=200
    Nth term is = 25
    common difference d = 50
    using equation a + (n-1)d
    200 +(25-1)50
    =1400 seats

    ReplyDelete
  2. sinMarch 18, 2009 at 2:08 PM

    First value=?
    d=6
    nth = 15
    ap = 250
    using equation a + (n-1)d, transpose.
    250 = a+(15-1)6
    a = 250 - 84
    a = 166

    ReplyDelete
  3. sinMarch 18, 2009 at 2:11 PM

    a + (n-1)d
    1st term = 5
    d=4
    nth = 23
    ap =

    5 + (23-1)4
    ap = 93

    ReplyDelete
  4. AnonymousMarch 20, 2009 at 9:10 AM

    Q4
    given a=5, d=3
    therefore the 15th term is =a+ 14d
    =5+ 14(3)= 5+ 42=47

    ReplyDelete
  5. AnonymousMarch 20, 2009 at 9:20 AM

    Q5
    using a+(n-1)d= nth term
    given d=2, nth term=15 a=?
    therefore we can say
    l=a+(n-1)d
    l=50 #0f items in d last term
    50=a+14(2)
    = 50-28=22

    ReplyDelete
  6. LykkeMarch 23, 2009 at 6:11 PM

    In arithmetic progression questions:
    a- 1st term
    d- common difference btwn terms
    n- number of terms/ elements
    to find the last term given all of the above=
    a+(n-1)d
    to find the sum of all terms given the values of a, d , and n you use the formula:
    >> n/2[2a+(n-1)d]
    ex.1
    a=200
    d=50
    n=25
    i.e the last term = a+(n-1)d
    = 200+(25-1)50
    = 1400

    ReplyDelete
  7. LykkeMarch 23, 2009 at 6:18 PM

    If a=?
    d=6
    n=15
    last term is 250
    you use the formula to find the last term and change the subject of the formula to get the first term a.
    last term= a+(n-1)d
    250= a+(15-1)6
    250= a+(14)6
    250= a+ 84
    therefore a= 250-84
    a= 166

    ReplyDelete
  8. LykkeMarch 23, 2009 at 6:23 PM

    given that n=23
    a=5
    d=4
    sum of all terms is = n/2[2a(n-1)d]
    = 23/2[2(5)+(23-1)4]
    = 11.5[10+(22)4]
    = 11.5[10+88]
    = 11.5[98]
    = 1127

    ReplyDelete
  9. DarkyMarch 31, 2009 at 6:29 AM

    First term a = 200
    commonm difference d = 50
    n term = 25
    a + (n-1)d
    200 + (25 -1 )50
    200 + 24 * 50
    1400
    There are 1400 seats in the 25Th row

    ReplyDelete
  10. DarkyMarch 31, 2009 at 6:45 AM

    common difference d = 6
    the first term a = ?
    nth term = 15

    ReplyDelete

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