SOLUTION: A well driller charges $9.00 per foot for the first 10 feet, $9.10 per foot for the next 10 feet, $9.20 per foot for the next 10 feet, and so on, at a price increase at $0.10 per f

Algebra ->  Sequences-and-series -> SOLUTION: A well driller charges $9.00 per foot for the first 10 feet, $9.10 per foot for the next 10 feet, $9.20 per foot for the next 10 feet, and so on, at a price increase at $0.10 per f      Log On


   



Question 1179299: A well driller charges $9.00 per foot for the first 10 feet, $9.10 per foot for the next 10 feet, $9.20 per foot for the next 10 feet, and so on, at a price increase at $0.10 per foot for succeeding intervals of 10 feet. How much does it cost to drill a well to a depth of 150 feet?
We need to solve this by using some sort of sequence and its equation. Thank you

Answer by ikleyn(52792) About Me  (Show Source):
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A well driller charges $9.00 per foot for the first 10 feet, $9.10 per foot for the next 10 feet,
$9.20 per foot for the next 10 feet, and so on, at a price increase at $0.10 per foot for succeeding
intervals of 10 feet. How much does it cost to drill a well to a depth of 150 feet?
We need to solve this by using some sort of sequence and its equation. Thank you
~~~~~~~~~~~~~~~~~

The depth of 150 ft is 10 times taken the depth of 15 feet.


So the person shoud pay $9.00  per first 15 ft;

                   then $9.10  per next  15 ft;

                   then $9.20  per next  15 ft;

 . . . . . . . . . . . . . . . . . . . . . . . . 

                   and so on until $10.40 per the last 15 ft.


Thus uou need to find the sum 


    9.00 + 9.10 + 9.20 + . . . + 10.40  dollars.


This sequense is an arithmetic progression with the first term equal to 9.00;

with the common difference of 0.10 and the number of terms 15.


There are different formulas and several ways to find the sum.


          The simplest way for beginner students is to find 
          the average value of the progression and multiply it 
          by the number of terms in the progression.


In this case, the average value of the progression is half the sum of the first and the last terms 

    %28a%5B1%5D%2Ba%5B15%5D%29%2F2 = %289.00+%2B+10.40%29%2F2 = 19.40%2F2 = 9.70  dollars;


So the total cost (the sum) is  9.70*15 = 145.50 dollars.


ANSWER.  $145.50.

Solved.

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For introductory lessons on arithmetic progressions see
    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.