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Mathematics 6 Online
OpenStudy (anonymous):

Given a geometric sequence in the table below, create the explicit formula and list any restrictions to the domain. n an 1 3 2 −6 3 12

OpenStudy (anonymous):

an = −2(3)^n − 1 where n ≥ 3 an = −3(3)^n − 1 where n ≥ 3 an = 3(−2)^n − 1 where n ≥ 1 an = 3(−3)^n − 1 where n ≥ 1

OpenStudy (anonymous):

@hartnn

hartnn (hartnn):

find the common ratio 'r' its is the ratio between consecutive terms

hartnn (hartnn):

r= 2nd term/ 1st term = 3rd term/ 2nd term = ... ??

OpenStudy (anonymous):

the only thing that i know is the formula an=ar^n−1

OpenStudy (anonymous):

i am pretty lost with this one so if you can explain

OpenStudy (anonymous):

it would be helpful

hartnn (hartnn):

r= 2nd term/ 1st term = 3rd term/ 2nd term did you understand this part ?

OpenStudy (anonymous):

what do you mean by 2nd term/ 1st term = 3rd term/ 2nd term??

hartnn (hartnn):

n an ----- n is the index, an is the n'th term 1 3 ----index 1, so 1st term is 3 2 −6 ------index 2 means 2nd term is -6 3 12 -------index 3 means 3rd term is 12

hartnn (hartnn):

now find the common ratio 'r' since you now know 1st , 2nd and 3rd terms

OpenStudy (anonymous):

\[\frac{ n_2 }{ n_1}\] and \[\frac{ n_3 }{ n_2 }\] Is what is meant Use the table to find the corresponding values of n

OpenStudy (anonymous):

oh ok im getting there so r would be -2

hartnn (hartnn):

thats correct! and we already have a = 1st term = 3 now just plug in values in the formula :)

OpenStudy (anonymous):

3(-2)^n-1

hartnn (hartnn):

\(\huge \checkmark \)

OpenStudy (anonymous):

thanks

hartnn (hartnn):

and n starts from 1 n = 1,2,3,4 ... so, \(\Large n \ge 1 \)

hartnn (hartnn):

welcome ^_^

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