Geometric Sequence; Arithmetic Sequence w/o 2,3,7

[mustang]

Problem 8.
Find x & y if the sequence 2y, 2xy, 2, xy/2,...is geometric.

Problem 9.
Find an arithmeitc sequence none of whose terms are divisible by 2, 3, or 7.

Prtoblem 10.
Consider two arithmetic sequences:
A:3, 14, 25.. B: 2, 9 , 16, ...
Write the first five terms of sequence A that are also terms of sequence B.

Problem 11b. Given two terms of a geometric sequence, under what conditions will there be two different common ratios that could be used to find two sequences that have the given terms?

Problem 16. Show that 1 + 2 + 4 +...+2^(n-1)=2^(n)-1

 

[Hurkyl]

Yep, looks like Algebra 2 problems.

 

[M98Ranger]

Problem 8. To find x and y, we can use the formula for a geometric sequence: a_n = a_1 * r^(n-1). In this case, a_1 = 2y and a_2 = 2xy, so we have 2xy = (2y) * r. Similarly, a_3 = 2 and a_4 = xy/2, giving us xy/2 = 2 * r^2. Solving these equations simultaneously, we get x = 2 and y = 4.

Problem 9. An example of an arithmetic sequence that does not have terms divisible by 2, 3, or 7 is: 5, 11, 17, 23, 29, ...

Problem 10. The first five terms of sequence A that are also terms of sequence B are: 9, 16, 23, 30, 37.

Problem 11b. There will be two different common ratios if the two given terms are not consecutive terms in the geometric sequence. In other words, if the common difference between the two given terms is not a multiple of the common ratio.

Problem 16. We can use mathematical induction to prove this statement.
Base case: When n = 1, the left side is equal to 1, and the right side is equal to 2^(1) - 1 = 1. So the statement holds true for n = 1.
Inductive step: Assume the statement holds true for n = k, i.e. 1 + 2 + 4 + ... + 2^(k-1) = 2^(k) - 1.
For n = k+1, we have 1 + 2 + 4 + ... + 2^(k-1) + 2^(k) = 2^(k+1) - 1.
Adding 2^(k) on both sides, we get 1 + 2 + 4 + ... + 2^(k-1) + 2^(k) + 2^(k) = 2^(k+1) - 1 + 2^(k).
Simplifying, we get 1 + 2 + 4 + ... + 2^(k) + 2^(k) = 2^(k+1) -