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As payment for his daughters yard work, a father agrees to give his daughter an allowance of $3.50 in the first week of the year with an increase of 50 cents each week until the last week of the year.

A) How much money did she receive for an allowance in the last week of the year?

B) What was the total amount of money her father gave her in allowances for the year?

I tried using a formula to solve this, and I thought it was the correct one, but I got a totally "way off" answer.

The formula I used was...

Sn = n[2a + (n-1)d]

The other formulas I learned about are:

Sn = n(a+tn)/2

tn = t1 + (n-1)d

I have no idea which one to use and how to solve. Really in need of help.

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4 answers
  1. Your first formula is incorrect , it should say
    Sn = (n/2)[2a + (n-1)d ]

    for a) you want to know what the value of term52 is
    so
    term52 = 3.5 + 51(.5)
    = 29

    you would use your first formula if you knew the first term, the common difference and the number of terms you have.
    That is the case in your problem b)
    a = 3.5 , d = .5, and n = 52
    so total amount of money
    = 26[7 + 51(.5)] = 845

    the second Sn formula can be used if you know the first and last terms.
    Since it is easier to use the second version, and we found the last term in a) we should get the same result
    Sum52 = 26( 3.5 + 29) = 845

    It should be easy to decide if you looking for the SUM of terms or if you are looking to find a particular TERM

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  2. Sorry I forgot part of the formula, it was: Sn = n[2a + (n-1)d]/2

    Where did you get 52 from? Did you divide 365 by 7? I got 52. something, didn't get the whole number 52.

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  3. mmmhhh.
    How many weeks are there in a year again?

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  4. 1 year = 52.177457 weeks

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