Might have to experiment a bit to choose the right answer.
In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.
Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456
In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.
Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.
Answer:

Step-by-step explanation:
Let x be the number of games
Cost of 1 game = $9
Cost of x games = 9x
he gets a student discount of $10.
So, Cost after discount = 9x-10
Let the total cost be T
So, 
So, An equation to represent the total cost of playing laser tag including the discount is 
The equation is:
h ( t ) = - 16 t² + 60 t + 3
The toy rocket will reach the ground when: h ( t ) = 0
- 16 t² + 60 t + 3 = 0

t 1 = - 0.05
t 2 = 3.8
Answer:
C ) 3.80 s