If the common ratio is r, then the sequence can be rewritten as 3, 3r, , . We see then that , which gives us that r=2. Therefore, the missing terms are 6 and 12.

If the common ratio is r, then the sequence can be rewritten as 3, 3r, , . We see then that , which gives us that r=2. Therefore, the missing terms are 6 and 12.

If the common ratio is r, then the sequence can be rewritten as 3, 3r, , . We see then that , which gives us that r=2. Therefore, the missing terms are 6 and 12.

The 4 consecutive of integers greater than 9 (but not including 9) are 11, 13, 15, 17. Added together, we get 56.

The 4 consecutive of integers greater than 9 (but not including 9) are 11, 13, 15, 17. Added together, we get 56.

The 4 consecutive of integers greater than 9 (but not including 9) are 11, 13, 15, 17. Added together, we get 56.

Because it is an arithmetic sequence, the difference between each term is the same. Therefore find out the difference between any two consecutive terms. and so the sequence increases by 5 each term. Thus the answer is .

Notice that between each of these numbers, there is a difference of . This means that for each element, you will add . The first element is or . The second is or , and so forth... Therefore, for the th element, the value will be or .

Notice that between each of these numbers, there is a difference of . This means that for each element, you will add . The first element is or . The second is or , and so forth... Therefore, for the th element, the value will be or .

Because it is an arithmetic sequence, the difference between each term is the same. Therefore find out the difference between any two consecutive terms. and so the sequence increases by 5 each term. Thus the answer is .