Answer:
First, let's define an arithmetic sequence:
In an arithmetic sequence, the difference between any two consecutive terms is always the same.
Then we can write it in a general way as:
aₙ = a₁ + (n - 1)*d
where:
aₙ is the n-th term of the sequence.
d is the constant difference between two consecutive terms.
a₁ is the initial term of our sequence.
Now in this case we know that the first terms of our sequence are:
84, 77, ...
Then we know the initial term of our sequence:
a₁ = 84.
And the value of d can be calculated as:
d = a₂ - a₁ = 77 - 84 = -7
Then the general way of writing this sequence is:
aₙ = 84 + (n - 1)*(-7)
And the recursion relation is:
aₙ = aₙ₋₁ - 7
So for the n-th term, we must subtract 7 of the previous term.
<h3>Question:</h3>
<em>Jon is selling tickets for the school talent show. On the 1st day, he sold 3 senior tickets and 12 child tickets for $195. On the 2nd day he sold 13 senior tickets for $299. Find the price of a senior citizen ticket.</em>
<h3>Answer:</h3>
Create a system of equations to help you solve this problem. The system of equations will look like: 3s + 12c = 195 and 13s = 299. The variable s represents the cost of senior tickets and the variable c represents the cost of children tickets.

Solve the second equation for the variable s as this is the easiest way to solve the problem. Solve the second equation for s by dividing both sides of the equation by 13 to isolate the variable s.
s = 23
Since the question was only asking for the price of a senior citizen ticket, you are technically done. The first equation was only put there to confuse you or allow you to check your work if you needed to. The price of a senior citizen ticket (variable s) is $23.
40 because 15×4 is 60 and there is 60 minutes in an hour
Answer:
$21
Step-by-step explanation:
First, find the price without the discount by adding 12 to 30:
30 + 12
= 42
To find the original cost of each game, divide 42 by 2
42/2
= 21
So, each game was originally $21