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Math Help - Solve Arithmetic progressison problem

  1. #1
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    Solve Arithmetic progressison problem

    1. Two jobs where offered to a man. In one the starting salary was $120,000 per month and the annual increment was $8,000. In the other job that salary commends at $85,000 per month but annual increment was $12,000. That man want to take a job that will give him more earning in the first twenty year, but was having problem of computation. Please advice him what to do.

    2. Find the sum of all the number between 200 and 400 that are divisible by 7
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by blenyo View Post
    1. Two jobs where offered to a man. In one the starting salary was $120,000 per month and the annual increment was $8,000. In the other job that salary commends at $85,000 per month but annual increment was $12,000. That man want to take a job that will give him more earning in the first twenty year, but was having problem of computation. Please advice him what to do.
    Hi blenyo,

    Nice jobs if you can get them.

    The annual salary for the first job is $1,440,000. The annual increment is $8,000.

    \sum\limits_{n=1}^{20}{1440000+(n-1)8000}

    The annual salary for the second job is $1,020,000. The annual increment is $12,000.

    \sum\limits_{n=1}^{20}{1020000+(n-1)12000}

    Use the formula for the sum of an arithmetic sequence

    S_n=\frac{n}{2}(a_1+a_n)

    to find the sums over 20 years and compare them.


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  3. #3
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    Quote Originally Posted by blenyo View Post
    1. Two jobs where offered to a man. In one the starting salary was $120,000 per month and the annual increment was $8,000. In the other job that salary commends at $85,000 per month but annual increment was $12,000. That man want to take a job that will give him more earning in the first twenty year, but was having problem of computation. Please advice him what to do.
    Done by masters.

    2. Find the sum of all the number between 200 and 400 that are divisible by 7
    7 divides into 200 28 times with remainder 4 which means that 203 is divisible by 7- 7 divides into 203 exacly 29 times. 7 divides into 400 57 times with remainder 1 which means that 399 is divisible by 7- 7 divides into 399 exactly 57 times. 57- 29= 28 so the number between 200 and 400 that are divisible by 7 are 203, 203+ 7, 203+ 2(7), ..., 203+ 28(7). That is an arithmetic progression which can be summed by the formula masters gave (and I'll bet it's in our textbook).
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