The local baseball team is advertising a special deal on ticket prices. The cost of the first ticket to a game is $12 and every ticket after that costs $0.25 less. If the maximum number of tickets you can buy is 30, what is the total cost of 30 tickets?
Wouldn't you use
Sn = n / 2 (a1 + an)
So you are finding
12 + 11.75 + 11.50 + ... for 30 terms
I would use
S(n) = n/2( 2a + (n-1)d) since we don't know the last term, as needed in your formula
I know we could find the last term, then use your formula but the above one would be faster in this case
S(30) = (30/2)(24 + 29(-.25))
= 15(16.75) = 251.25
Actually, for this particular question, we can use a different formula called the arithmetic sequence formula.
In this case, the first ticket costs $12, and the cost decreases by $0.25 for every subsequent ticket. This means that the common difference between the ticket prices is -$0.25.
To find the total cost of purchasing 30 tickets, we can use the arithmetic sequence formula:
Sn = (n/2) * (2a1 + (n-1)d)
Where:
Sn is the sum of the arithmetic sequence
n is the number of terms (tickets) in the sequence
a1 is the first term in the sequence
d is the common difference
Plugging in the values:
n = 30 (maximum number of tickets)
a1 = 12 (cost of the first ticket)
d = -0.25 (common difference)
Sn = (30/2) * (2*12 + (30-1)*(-0.25))
= 15 * (24 + 29*(-0.25))
= 15 * (24 - 7.25)
= 15 * 16.75
= $251.25
Therefore, the total cost of 30 tickets is $251.25.
Yes, you can use the formula for the sum of an arithmetic series, also known as the arithmetic progression, to solve this problem.
The formula you provided, Sn = (n / 2) * (a1 + an), is used to find the sum (Sn) of an arithmetic series with a certain number of terms (n), where a1 represents the first term and an represents the last term.
However, in this case, the cost of the tickets decreases with each subsequent ticket, so it does not form an arithmetic series with a constant common difference. Instead, we need to approach this problem by using a different method.
To find the total cost of 30 tickets, we can calculate it by adding up the cost of each individual ticket.
Here's how you can calculate it:
1. Determine the cost of the first ticket: $12.
2. Determine the cost of each subsequent ticket: The cost of every ticket after the first one decreases by $0.25. So, the second ticket would cost $12 - $0.25, the third ticket would cost $12 - 2 * $0.25, and so on.
3. Calculate the total cost: Add up the costs of all 30 tickets.
Let's calculate it step by step:
Cost of the first ticket = $12.
Cost of the second ticket = $12 - $0.25 = $11.75.
Cost of the third ticket = $12 - 2 * $0.25 = $12 - $0.50 = $11.50.
We can see that with each subsequent ticket, the cost decreases by $0.25.
Continuing this pattern, we can calculate the cost of the 30th ticket:
Cost of the 30th ticket = $12 - 29 * $0.25.
Now, let's calculate the total cost by adding up the costs of all 30 tickets:
Total cost = Cost of the first ticket + Cost of the second ticket + ... + Cost of the 30th ticket.
Total cost = $12 + ($12 - $0.25) + ($12 - 2 * $0.25) + ... + ($12 - 29 * $0.25).
Total cost = 30 * $12 - ($0.25 + 2 * $0.25 + ... + 29 * $0.25).
Total cost = 30 * $12 - $0.25 * (1 + 2 + ... + 29).
To calculate the sum of the numbers from 1 to 29, we can use the formula for the sum of an arithmetic series.
Sum of numbers from 1 to 29 = (n / 2) * (a1 + an) = (29 / 2) * (1 + 29) = 15 * 30 = 450.
Plugging in this value, we have:
Total cost = 30 * $12 - $0.25 * 450.
By performing this calculation, you can find the total cost of the 30 tickets.