Question

Which sequence is modeled by the graph below?
(Geometric sequencing!!!!) Fan/Medal

Answer 1

Yes, it a geometric sequence. where x-coordinate represents terms location and y-coordinate represents the value.

Sequence follows;
5, 0.5, 0.05 ....

To confirm use the sequence formula;\[S_{n}=\frac{ a_{1}(1-r^{n}) }{ (1-r) }\]where r= common ratio, a= first term, n= first "n" terms.

Answer 2

You can see that the sequence have common ratio;
\[\frac{ 0.5 }{ 5 }=0.1\]and\[\frac{ 0.05 }{ 0.5 }=0.1\]

Answer 3

Please use the equation editor in the bottom.

Do you mean?
\[a_{n} = 3(1)^{n-1} \] ...

Answer 4

Seems like the options are incorrect, there is no way you can get "5" using "n=1"

Answer 5

oh im sorry I was looking at the same problem but different problem

an = 5(−10)^(n − 1)

an = 0.5(10)^(n − 1)

an = one tenth(5)^(n − 1)

an = 5(one tenth)^(n − 1)

Answer 6

\[ok\]

Answer 7

Correct, last option is right.

Answer 8

sweet ok now ummm

Answer 9

I think its a_n = 3(1)^(n − 1)

Answer 10

remember "1^anything" will always be "1" so it can not be a correct answer and please post all the options.

Answer 11

a_n = 3(1)^(n − 1)

a_n = 1 + (3)^(n − 1)

a_n = (3)(n − 1)

a_n = 1(3)^(n − 1)

Answer 12

what you think is the correct option now?

Answer 13

All you need to do is put the x-coordinate in place of "n" and confirm the y-coordinate (the answer)

Answer 14

\[3(1)^{1-1}=3\] but at x=1, y=1 so its incorrect.
\[1+(3)^{1−1}=1+1=2\]which is also incorrect.
\[3(n-1)=3(1-1)=0\]Last one;\[(1)(3)^{1-1}=1\] seems to be correct, lets check the second term by putting "n=2" where "x=2", "y=3"\[(1)(3)^{2-1}=(1)(3)^{1}=3\]

Answer 15

So which option is correct?

Answer 16

the last one? its the only one that would make sense in my opinion

Answer 17

Yes, the last one is the correct answer.

Answer 18

thank you :)