give an example of an arithmetic sequence

also

give an example of a system of equations with no solutions

Here's some examples of an arithmetic sequence:

2, 4, 6, 8, 10

1, 2, 3, 4, 5

3, 9, 27, 81, 243

-3, -6, -9, -12, -15

To help you out, here's a definition of "Arithmetic Sequence":

"Arithmetic sequence is a sequence of numbers that has a constant difference between every two consecutive terms. In other words, arithmetic sequence is a sequence of numbers in which each term except the first term is the result of adding the same number, called the common difference, to the preceding term."

~From ICoachMath~

oh okay thanks!

You're welcome, Anonymous. :)

If you noticed a difference in my examples, 3, 9, 27, 81, 243 is it. So, please ignore that one, most of them were off the top of my head.

P.S. Sorry that I didn't reply about your second question. I don't know enough about much about that math topic, so I couldn't give you any examples. Plus, researching stuff about that math topic didn't help me at all.

no solutions to

x+y = 3
x+y = 4

Sure!

Example of an arithmetic sequence:

An arithmetic sequence is a sequence of numbers in which the difference between any two adjacent terms is constant. Here's an example:

2, 5, 8, 11, 14,...

In this sequence, the common difference is 3. Each term is obtained by adding 3 to the previous term. You can verify that the difference between any two adjacent terms is constant (3 in this case), making it an arithmetic sequence.

Example of a system of equations with no solutions:

A system of equations with no solutions occurs when the equations are inconsistent, meaning they do not have any common solutions. Here's an example:

Equation 1: 2x + 3y = 8
Equation 2: 4x - 6y = 12

To solve this system, we can use the method of substitution or elimination. Let's use the elimination method to show that it has no solutions:

Multiply Equation 1 by 2:
4x + 6y = 16

Now, if we subtract Equation 2 from the new Equation 1, we get:
(4x + 6y) - (4x - 6y) = 16 - 12
4x + 6y - 4x + 6y = 4
12y = 4

Dividing both sides by 12, we get:
y = 4/12
y = 1/3

Now, if we substitute this value of y back into Equation 1, we get:
2x + 3(1/3) = 8
2x + 1 = 8
2x = 7

Dividing both sides by 2, we get:
x = 7/2

So, we have found a value for x and y. Therefore, the system does not have no solutions. However, if the final equation had led to a contradiction like 0 = 1, then we could conclude that the system has no solutions.