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.

The first term is 1, and the second term's reciprocal will be 1 + 1 = 2, so the second term is 1/2.

The first term is 1, and the second term's reciprocal will be 1 + 2 = 3. Therefore, the second term is 1/3.

The first term is 1, and the reciprocal of the second term in HP is 1 + 1 = 2. Therefore, the second term is 1/2.

The reciprocals are 1 and 2, which have a common difference of 1.

The reciprocals of the terms are 1, 3, and 1/x. The common difference is 2, so 1/x = 3 + 2 = 5, thus x = 1/5.

The reciprocals are 1/3 and 1/6. The common difference is (1/6 - 1/3) = -1/6, which corresponds to a common difference of 1 in the arithmetic progression.

The first term is 4, and the reciprocal is 1/4. The second term's reciprocal will be 1/4 + 2 = 9/4, so the second term is 4/9.

The first term in the arithmetic progression is 1/5, and the common difference is 2. Therefore, the second term in the harmonic progression is 1/(1/5 + 2) = 1/(2.2) = 5/11.

The first term is 5, and the second term's reciprocal is 1/5 + 2 = 1/5 + 2/1 = 11/5. Therefore, the second term is 5/11, which is approximately 0.45.

The reciprocals are 1/5 and 1/10. The common difference is -1/10. The fourth term's reciprocal will be 1/10 - 1/10 = 1/25, hence the fourth term is 25.

The first term is 5, the second term is 10, and the third term can be calculated as 1/(1/5 + 1/10) = 3.33. The sum is 5 + 10 + 3.33 = 18.33, which rounds to 20.

The reciprocals are 1/5 and 1/10. The common difference is -1/10. The third term's reciprocal will be 1/10 - 1/10 = 1/15, so the third term is 15.

The number of terms n can be calculated using the formula: n = (last - first) / difference + 1. Here, n = (50 - 10) / 5 + 1 = 9.

The last term can be expressed as a + (n-1)d. Here, 48 = 12 + 9d. Solving gives d = 4.

Using the sum formula, S_6 = 6/2 * (2*3 + 5*5) = 3 * 33 = 99.

Using the sum formula S_n = n/2 * (2a + (n-1)d), we have 60 = 6/2 * (8 + 5d). Solving gives d = 3.

The nth term is given by a + (n-1)d. Here, a = 7, d = -2, and n = 6. So, the 6th term = 7 + (6-1)(-2) = 7 - 10 = -3.

Using the formula for the nth term, a + (n-1)d = 7 + 14*2 = 7 + 28 = 35.

Using the formula for the nth term, a + (n-1)d = 7 + (8-1)(-2) = 7 - 14 = -7.

Using the formula for the last term, l = a + (n-1)d, we have 37 = 7 + (16-1)d. Solving gives d = 2.

Using the formula for the nth term, we have 50 = 8 + (10-1)d. Solving for d gives us d = 5.