Question

Tell whether the sequence 1/3,0,1,-2… is arithmetic, geometric, or neither.

Answer 1

The answer is D. hope that helps

Answer 2

Answer:

A. neither; 7, -20, 61  

Step-by-step explanation:

Let's remember the definitions of arithmetic and geometric sequences:

$\fbox {Arithmetic sequence:}$ the difference between one term and the next is a constant (always the same), i.e. we add the same value to one term to get the next one.

$\fbox {Geometric sequence:}$each term is found by multiplying the previous term by a constant.

Let's see that the sequence is neither of both:

We have to add (-1/3) to the first term to get the second one. But if we add (-1/3) to the second term, we get -1/3 and that's not our third term.

It can't be an arithmetic sequence.

Something similar happens when we want to see if it's a geometric sequence:

We have to multiply the first term by 0 to get the second one. But if we multiply the second term by 0, we still get 0, and that's not our third term.

It can't be a geometric sequence.

At this point is clear that the correct option is A. because is the only one with the option "neither", but let's find the next three terms of the sequence:

Let's notice that from the first term to the second we subtract 1/3.

From the second term to the third, we add 1.

From the third to the fourth, we subtract 3.

In general, to get the n term, we add $(-1) ^(n-1) *3^(n-3)$ to the previous term

The $(-1) ^(n-1)$ part is because we are adding a positive number when n is odd and a negative when n is even.

Let's see that this expression matches what we have:

n = 2, $1/3 + (-1) ^ (2-1)*3^ (2-3) = 1/3 + (-1) ^1 * 3^( -1) = 1/3 - 3^ (-1) = 1/3 - 1/3$ = 0

n= 3, $0 + (-1) ^ (3-1) *3^ (3-3) = 0 + (-1) ^2 * 3^0 = 0 + 1 =$ 1

n = 4, $1 + (-1) ^ (4-1) *3^ (4-3) = 1 - 3^1 =$ -2

So, the next three terms are

n = 5, $-2 + (-1) ^ (5-1) *3^ (5-3) = -2 + 3^2 =$ 7

n = 6, $7 + (-1) ^ (6-1) *3^ (6-3) = 7 - 3^3=$ -20

n = 7, $-20 + (-1) ^ (7-1) *3^ (7-3) = -20 + 3^4 =$ 61