Question

four students wrote sequences in math class. angela -6,-9,-12,-15,... bradley -2,-6, -12,-24,... carter -1, -3, -9, -27,... dominique -1, -3,-9, -81,... which student wrote a geometric sequence? a)angela b)bradley c)carter d)dominique

Answer 1

Answer

c) carter

Explanation

Remember that A geometric sequence is a sequence where each term after the first is found by multiplying the previous term by a fixed number called the ratio. So to prove if a sequence is geometric, y we just need to verify that the ratio between any tow terms of the sequence is the same.

Now, to find the ratio of a geometric sequence, we use the formula:

$r=\frac{a_{n}}{a_{n-1}}$

$r$ is the ratio

$a_{n}$ is the current term

$a_{n-1}$ is the previous term

The formula just mean that we can find the ratio by dividing the current term and the previous term.

Let's apply this to our sequences:

Angela -6,-9,-12,-15,...

Take -9 as the current term and -6 as the previous term, so $a_{n}=-9$ and $a_{n-1}=-6$; now let's find the ratio.

$r=\frac{a_{n}}{a_{n-1}}$

$r=\frac{-9}{-6} =\frac{3}{2}$

Lets check if the ratio hold for the next pair -9 and -12. $a_{n}=-12$ and $a_{n-1}=-9$, so

$r=\frac{-12}{-9} =\frac{4}{3}$

The ratio is not the same, so Angela's sequence is not a geometric one.

Bradley -2,-6, -12,-24,...

For -2 and -6

$r=\frac{a_{n}}{a_{n-1}}$

$r=\frac{-6}{-2} =3$

For -6 and -12

$r=\frac{-12}{-6} =2$

The ratio is not the same, so Bradley's sequence is not a geometric one.

Carter -1, -3, -9, -27,...

For -1 and -3

$r=\frac{-3}{-1} =3$

For -3 and -9

$r=\frac{-9}{-3} =3$

For -9 and -27

$r=\frac{-27}{-9} =3$

The ratio is always the same, so Carter's sequence is a geometric one.

Dominique -1, -3,-9, -81,...

For -1 and -3

$r=\frac{-3}{-1} =3$

For -3 and -9

$r=\frac{-9}{-3} =3$

For -9 and -81

$r=\frac{-81}{-9} =9$

The ratio is not the same, so Dominique's sequence is not a geometric one.

Answer 2

Answer:

C. carter

Step-by-step explanation: