The intersection of sets A and B is the set of elements that are common to both sets, which is {4, 5}.

The multiples of 3 less than 30 are {3, 6, 9, 12, 15, 18, 21, 24, 27}, totaling 9 elements.

The difference P - Q includes elements in P that are not in Q, which is {1, 2}.

Set P = {2, 4, 6, 8} and set Q = {2, 3, 5, 7}. The union is {2, 3, 4, 5, 6, 7, 8}.

The intersection of sets P and Q includes elements that are both even and prime. The only even prime number is 2.

Set P = {2, 4, 6, 8} and set Q = {2, 3, 5, 7}. The difference P - Q = {4, 6, 8}.

The intersection P ∩ Q includes only the even prime number, which is {2}.

The symmetric difference of sets R and S includes elements that are in either set but not in both. Thus, it is {1, 2, 3, 6, 7}.

The difference R - S includes elements in R that are not in S, which is {1, 2}.

The symmetric difference is the set of elements in either set R or set S but not in both, which is {1, 2, 5, 6}.

The union of sets X and Y includes all unique elements from both sets. Thus, the union is {a, b, c, d}.

The sum of the first n terms S_n = n/2 * (2a + (n-1)d). Here, S_10 = 10/2 * (2*4 + 9*3) = 5 * (8 + 27) = 5 * 35 = 175.

Let the first term be a and the common ratio be r. Then, 2nd term = ar = 8 and 4th term = ar^3 = 32. Dividing these gives r^2 = 4, so r = 2.

Let the first term be a and the common ratio be r. Then, 2nd term = ar = 12 and 4th term = ar^3 = 48. Dividing these gives r^2 = 4, so r = 2.

Let the first term be a and the common ratio be r. Then, 2nd term = ar = 8 and 4th term = ar^3 = 32. Dividing gives r^2 = 4, so r = 2.

Let the first term be a and the common difference be d. From the equations a + d = 10 and a + 4d = 16, we can solve for d to find it is 2.

Let the first term be a and the common difference be d. From a + d = 10 and a + 4d = 16, we can find a + 2d = 12, which is the 3rd term.

Let the first term be a and the common difference be d. From the equations a + d = 15 and a + 3d = 25, we can find d = 5.

Let the first term be a and the common difference be d. From the equations a + d = 8 and a + 3d = 14, we can find a = 6.

Let the first term be a and the common difference be d. From the equations a + d = 8 and a + 4d = 14, we can find the 3rd term a + 2d = 10.

Let the first term be a and the common difference be d. From the equations a + d = 8 and a + 4d = 20, we can solve for a to find it equals 4.

The 3rd term is given by ar^2. So, 12 = a(2^2) => a = 12/4 = 3.

Let the first term be a. Then, the 3rd term is ar^2 = 27. Thus, a * 3^2 = 27, giving a = 3.

Let the first term be a and the common difference be d. We have a + 2d = 12 and a + 6d = 24. Subtracting these gives 4d = 12, so d = 3.

Let the first term be a and the common difference be d. From the equations a + 2d = 15 and a + 5d = 24, solving gives d = 3.

Let the first term be a and the common difference be d. From the equations a + 2d = 15 and a + 6d = 27, solving gives d = 3.

Let the first term be a and the common difference be d. Then, a + 2d = 12 and a + 6d = 24. Solving these gives d = 3.

Let the first term be a and the common difference be d. From the equations a + 4d = 15 and a + 9d = 30, we can find d = 3.

Let the first term be a and the common difference be d. From the equations a + 4d = 20 and a + 9d = 35, we can solve for a to find it is 10.

Let the first term be a and the common difference be d. From the equations a + 5d = 30 and a + 8d = 45, we can find d = 5.