Question 1198920
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x = amount deposited in January


Since the deposits increase by the same amount ($1.75) each month, we have an arithmetic sequence


a1 = first term = x
d = common difference = 1.75
n = number of terms


We'll have n = 12 to represent the 12 months.
Let's find the sum of the first 12 terms of this sequence.
This will get the total amount saved after December's deposit.


Sn = sum of the first n terms of an arithmetic sequence
Sn = (n/2)*(2*a1+d*(n-1))
S12 = (12/2)*(2*x+1.75*(12-1))
S12 = 12x+115.5



The sum of the first 12 terms of the sequence {x, x+1.75, (x+1.75)+1.75, etc} is 12x+115.5


Set this equal to 157.50 which is the total amount saved at the end of December.


Solve for x.
12x+115.5 = 157.50
12x = 157.50-115.5
12x = 42
x = 42/12
x = 3.5


You deposited $3.50 into the piggybank in January to start things off.
Then for February, the deposit is 3.50+1.75 = 5.25 and so on until you arrive at December.


I don't recommend writing out all the terms as the tutor @greenestamps has done, if doing it by hand. 
Though you can use computer software to do it for you. A spreadsheet is really handy in situations like this.


Let's return back to the summation formula mentioned earlier.
This time we plug in n = 4 to determine how much is saved after April's deposit.
I.e. this is the sum of the first four terms.


Sn = (n/2)*(2*a1+d*(n-1))
S4 = (4/2)*(2*3.5+1.75*(4-1))
S4 = 24.5


Answer: <font color=red>$24.50</font>
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