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

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 third term's reciprocal will be 1/10 - 1/10 = 1/15, so the third term is 15.

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 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.

The nth term of an arithmetic sequence is given by a + (n-1)d. Here, a = 5, d = 3, n = 10. So, 5 + (10-1)3 = 32.

The reciprocals are 1, 2, and 3. The common difference is 2 - 1 = 1.

The reciprocals are 1, 2, and 3, which are in arithmetic progression. The next term in the sequence of reciprocals is 4, so the fourth term is 1/4.

The reciprocals of the terms form an arithmetic progression: 2, 3, and x. The common difference is 1. Therefore, x = 6.

In a harmonic progression, the reciprocals of the terms form an arithmetic progression, hence 1/a + 1/b = 1/c.

In a harmonic progression, the reciprocals of the terms form an arithmetic progression, which leads to the relationship 1/a + 1/c = 2/b.

In a harmonic progression, the reciprocals of the terms are in arithmetic progression, hence 1/a, 1/b, 1/c are in AP.

Substituting x = 3 into the function gives f(3) = 2(3) + 1 = 6 + 1 = 7.

The transformation g(x) = 2(x - 1) + 3 indicates a horizontal shift to the right by 1 unit.

If the GCD of two numbers is 1, it means they have no common factors other than 1, thus they are coprime.