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    Candicane's Avatar
    Candicane Posts: 3, Reputation: 1
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    #1

    Aug 23, 2012, 10:10 AM
    Story problem
    A person hired a firm to build a CB radio tower. The firm charges $100 for labor for the first 10 feet. After that, the cost of the labor for each succeeding 10 feet is $25 more than the preceding 10 feet. That is, the next 10 feet will cost $125, the next 10 feet will cost $150, etc. How much will it cost to build a 90-foot tower?

    I just can't figure out what fomula to use!
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Aug 23, 2012, 10:27 AM
    Do it bit by bit if you can't make a formula. You might find a pattern, and eventually the formula.
    Candicane's Avatar
    Candicane Posts: 3, Reputation: 1
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    #3

    Aug 23, 2012, 11:13 AM
    Quote Originally Posted by Unknown008 View Post
    Do it bit by bit if you can't make a formula. You might find a pattern, and eventually the formula.
    Thanks I just wrote 10, 9 times and added 100 and the added the twenty five to it all the way down the line I got 300. It just doesn't seem right to me!
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Aug 23, 2012, 11:17 AM
    Think of it in terms of sections. There are a total of 9 sections, and if we number them 1 through 9 the cost to build the ith section is 75+25i. The formula for summing this up for all 9 sections is:



    This sum can be broken up into:



    All you need now is to remember that

    ,

    and



    Hope this helps.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Aug 23, 2012, 11:37 AM
    A person hired a firm to build a CB radio tower. The firm charges $100 for labor for the first 10 feet. After that, the cost of the labor for each succeeding 10 feet is $25 more than the preceding 10 feet. That is, the next 10 feet will cost $125, the next 10 feet will cost $150, etc.

    How much will it cost to build a 90-foot tower?
    Doing it the 'manual' way:
    First 10 feet = $100 (remaining 80 feet)
    Second 10 feet = $125 (remaining 70 feet)
    Third 10 feet = $150 (remaining 60 feet)
    Forth 10 feet = $175 (remaining 50 feet)

    It's kind of a pattern here, something like:
    100, 125, 150, 175, 200,.

    This is an arithmetic progression with first term a = 100, common difference d = 25 and with n = 9 terms.

    And you should have the formula to get the sum of the first 9 terms of the sequence.

    EDIT: Which is another way of thinking about it as compared to ebaines' answer above :)
    Candicane's Avatar
    Candicane Posts: 3, Reputation: 1
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    #6

    Aug 23, 2012, 11:57 AM
    Quote Originally Posted by Unknown008 View Post
    Doing it the 'manual' way:
    First 10 feet = $100 (remaining 80 feet)
    Second 10 feet = $125 (remaining 70 feet)
    Third 10 feet = $150 (remaining 60 feet)
    Forth 10 feet = $175 (remaining 50 feet)

    It's kind of a pattern here, something like:
    100, 125, 150, 175, 200, ...

    This is an arithmetic progression with first term a = 100, common difference d = 25 and with n = 9 terms.

    And you should have the formula to get the sum of the first 9 terms of the sequence.

    EDIT: Which is another way of thinking about it as compared to ebaines' answer above :)
    Thanks you so much, I had the pattern, just could plug it into a fomula
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #7

    Aug 23, 2012, 12:02 PM
    The formula is really just like an average formula if you examine it well.

    It's the number of terms, multiplied by the first and the last term, divide by 2.

    1, 2, 3, 4, 5

    To get the sum, you would make everything 3's and multiply by 5.

    3, 3, 3, 3, 3

    In other words, you took 1 from 4 and gave it to 2, took 2 from 5 to give it to 1.

    So, it really becomes the average of the first and last terms ((5+1)/2) = 3, times the total number of terms, 5.



    the first term is 'a'.

    The last term is given by T_n = a + (n-1)d [and this would be the only formula you cannot afford to forget in arithmetic progression]

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