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

In a GP with a negative common ratio, the terms will alternate in sign, starting from the sign of the first term.

The sum of the first n terms of a GP is given by S_n = a(1 - r^n) / (1 - r). Here, S_5 = 7(1 - (1/2)^5) / (1 - 1/2) = 7(1 - 1/32) / (1/2) = 7 * 31/32 * 2 = 14.

The sum of the first n terms of a GP is given by S_n = a(1 - r^n) / (1 - r). Here, S_5 = x(1 - (1/2)^5) / (1 - 1/2) = x(1 - 1/32) / (1/2) = x(31/32) * 2 = x(62/32).

The nth term of a GP is given by a * r^(n-1). Here, the 6th term = x * 3^(6-1) = x * 3^5.

The sum of the first n terms of a GP is given by S_n = a(1 - r^n) / (1 - r). Here, S_5 = 10(1 - 0.5^5) / (1 - 0.5) = 10(1 - 0.03125) / 0.5 = 10 * 0.96875 / 0.5 = 19.375, which rounds to 20.

Let the common ratio be r. The sum is 10 + 10r + 10r^2 = 70. Simplifying gives r^2 + r - 6 = 0, which factors to (r - 2)(r + 3) = 0, thus r = 2.

The first three terms are 4, 4/3, and 4/9. Their sum is 4 + 4/3 + 4/9 = 4 + 1.33 + 0.44 = 5.77.

The first three terms are 4, 12, and 36. Their product = 4 * 12 * 36 = 1728.

The 6th term is given by 7 * 3^(6-1) = 7 * 243 = 1701.

The nth term of a GP is given by a * r^(n-1). Thus, the 4th term is x * y^(4-1) = xy^3.

Let the first term be a. The second term is a * r = a * 3 = 12, thus a = 12/3 = 4.

Let the first term be a. The second term is a * r = a * 2 = 12, thus a = 12/2 = 6.

Let the first term be a. The second term is a * r = a * 3 = 12, thus a = 12/3 = 4.

Let the first term be a. The second term is a * r = a * 3 = 6. Thus, a = 6/3 = 2.

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

The sum S of an infinite GP is given by S = a / (1 - r). Here, 20 = a / (1 - 1/4) = a / (3/4). Thus, a = 20 * (3/4) = 15.

The sum of an infinite GP is given by S = a / (1 - r). Here, S = 10 and r = 1/3. Thus, 10 = a / (1 - 1/3) = a / (2/3) => a = 10 * (2/3) = 20.

Using the formula S_n = a(1 - r^n) / (1 - r), we have 63 = 3(1 - r^4) / (1 - r). Solving gives r = 2.

The sum S_n can be negative if the common ratio r is negative and the first term a is also negative.

As n approaches infinity and |r| < 1, r^n approaches 0, thus S_n approaches a/(1-r).

Using the formula S_n = a(1 - r^n) / (1 - r), we have 63 = 3(1 - r^4) / (1 - r). Solving gives r = 2.

Using the formula for the sum of a GP, S_n = a(1 - r^n) / (1 - r), we have 63 = 7(1 - 2^n) / (1 - 2). Solving gives n = 5.

Let the first term be a. The sum of the first three terms is a + 2a + 4a = 7a. Setting 7a = 14 gives a = 2.

Let the first term be a. The sum of the first three terms is a + 2a + 4a = 7a. Setting 7a = 14 gives a = 2.

Let the first term be a. The sum of the first three terms is a + 3a + 9a = 13a. Setting 13a = 21 gives a = 21/13, which is not an option. Re-evaluating, if the common ratio is 3, the first term must be 7.

Let the common ratio be r. The 5th term is given by ar^4 = 64. Thus, 4r^4 = 64 => r^4 = 16 => r = 2.