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anonymous
 5 years ago
What is the next number in the sequence: 9, 3, 1, 1/3,...?.....I have alot of these questions with different sequence numbers how do I determine the the squence for each without spending alot of time on one question?
anonymous
 5 years ago
What is the next number in the sequence: 9, 3, 1, 1/3,...?.....I have alot of these questions with different sequence numbers how do I determine the the squence for each without spending alot of time on one question?

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heisenberg
 5 years ago
Best ResponseYou've already chosen the best response.0These types of questions are all about determining patterns between the numbers. There is no "goto" method to doing so, but a good start is to compare each number with the previous number and see any relations.

heisenberg
 5 years ago
Best ResponseYou've already chosen the best response.0here we can see that it goes: 9, 3... already i know that 9 / 3 = 3, and also \[\sqrt(9) = 3\]

heisenberg
 5 years ago
Best ResponseYou've already chosen the best response.0moving on, I see: 3, 1... well the sqrt(3) != 1, but 3 / 3 = 1

heisenberg
 5 years ago
Best ResponseYou've already chosen the best response.0(!= means does not equal)

heisenberg
 5 years ago
Best ResponseYou've already chosen the best response.0I'm starting to see that each number is just the previous divided by 3. the remaining numbers confirm this. so if i have: \[\frac{1}{3} / 3 = \frac{1}{3} * \frac{1}{3} = \frac{1}{9}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in problems like this one how will I know to add, subtract, divide..etc to figure out how the sequence is going without having to use every number to get to the right one

heisenberg
 5 years ago
Best ResponseYou've already chosen the best response.0You'll never really know which to use, but the idea is to eyeball the numbers and play around. The whole goal is to try and detect patterns. Like I did up there, examine the relationship between two consecutive numbers. Then take any of those that worked and try them on another pair.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok....very helpful thank you!
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