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.

Question

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.

Answer 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

Answer 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.

Answer 3

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

Answer 4

1 year = 52.177457 weeks