The last term of an AP is given by a + (n-1)d. Here, 100 = 10 + (20-1)d. Solving gives d = 5.

The common difference d can be calculated using the formula for the nth term: a + (n-1)d = last term. Here, 12 + 9d = 48, solving gives d = 4.

The sum of the first n terms S_n = n/2 * (2a + (n-1)d). Here, S_10 = 10/2 * (2*7 + 9*2) = 5 * (14 + 18) = 5 * 32 = 160.

The nth term is given by a + (n-1)d. Here, a = 2, d = 3, and n = 15. So, the 15th term = 2 + (15-1) * 3 = 2 + 42 = 44.

The first term a = 2 and the common difference d = 3. The 15th term = a + (15-1)d = 2 + 42 = 44.

The common difference d can be found using the formula for the nth term. The last term is given by a + (n-1)d. Here, 48 = 12 + (8-1)d, solving gives d = 4.

Using the formula for the last term: a + (n-1)d = last term, we have 8 + 5d = 32. Solving gives d = 4.

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 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.

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. 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.

Using the formula for the nth term, a + 6d = 25. Substituting d = 3 gives a + 18 = 25, thus a = 7.

Using the formula for the nth term, a + 6d = 50. Substituting d = 5 gives a + 30 = 50, hence a = 20.

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.

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

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

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 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.