|
Question 1198920: I'm saving money over a year in order to buy a $157.50 gift. I start with an empty piggy bank. In January I put in a small amount of money. Each month after January, I put in $1.75 more than I put in the previous month. By the end of December, I've saved $157.50 exactly. The amount in dollars that I had in my piggy bank after April's deposit was
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The amounts put in the piggy bank in each month form an arithmetic sequence with a common difference of 1.75.
The sum of the terms of an arithmetic sequence is
(number of terms) times (average of the terms)
= (number of terms) times (average of first and last terms)
= (number of terms) times ((first plus last)/2)
= (half the number of terms) times (first plus last)
The sum of the 12 terms for the 12 months is the final amount, 157.50, so
6(first plus last) = 157.50
first plus last = 157.50/6 = 26.25
The sum of the 6th and 7th terms is the same as the sum of the first and last; and the difference between the 6th and 7th terms is 1.75. Let x and y be the amounts put in the piggy bank in the 6th and 7th months; then
y+x=26.25
y-x=1.75
2y=28
y=14
The amount put in the piggy bank in the 7th month (July) was $14.
The amount put in the piggy bank in the first month was $14, minus the common difference of $1.75 6 times: $14-6($1.75) = $14-$10.50 = $3.50.
The amounts put in the piggy bank in each month were then
$3.50, $5.25, $7, $8.75, $10.50, $12.25, $14, $15.75, $17.50, $19.25, $21, and $22.75
The amount in the piggy bank after April's deposit was
ANSWER: $3.50 + $5.25 + $7 + $8.75 = $24.50
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
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: $24.50
|
|
|
| |
