The sum S_5 = 5/2 * (2a + 4d). Here, 50 = 5/2 * (2a + 8), solving gives a = 10.

The average of the first n terms is given by S_n/n. Here, S_5 = 50, so the average = 50/5 = 10.

The sum of the first n terms S_n = n/2 * (2a + (n-1)d). By differentiating S_n with respect to n, we can find the common difference. The common difference is 3.

The common difference can be found by calculating S_n - S_(n-1). S_n = 3n^2 + 2n and S_(n-1) = 3(n-1)^2 + 2(n-1). Simplifying gives the common difference as 6.

The nth term of an AP is given by the formula a + (n-1)d. Here, a = 5, d = 3, and n = 10. Thus, the 10th term = 5 + (10-1) * 3 = 5 + 27 = 32.

Let the common difference be d. From the equations 4 + 5d = 24, solving gives d = 4.

The 6th term is given by a + 5d. Here, it is x + 5*2 = x + 10.

Let the first term be a and the common difference be d. From the equations a + d = 10 and a + 3d = 14, we can solve for a and find it to be 8.

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

Let the first term be a and the common difference be d. From the equations a + 2d = 12 and a + 6d = 24, we can find d = 3.

Let the first term be a and the common difference be d. From the equations a + 2d = 15 and a + 5d = 24, we can solve for d, which gives d = 3.

Let the first term be a and the common difference be d. From the equations a + 3d = 20 and a + 6d = 26, we can solve for a and find it to be 12.

Let the first term be a and the common difference be d. From the given terms, we have a + 4d = 20 and a + 9d = 35. Solving these gives a = 10.

The sum of the first n terms is given by S_n = n/2 * (2a + (n-1)d). Here, S_8 = 8/2 * (20 + 35) = 4 * 55 = 220.

Using the formula for the last term, 48 = 12 + (10-1)d. Solving gives d = 4.

Using the nth term formula, we have 40 = 4 + (10-1)d. Solving gives d = 4.

The nth term of an AP is given by a + (n-1)d. Here, a = 5, d = 3, and n = 10. So, the 10th term = 5 + (10-1) * 3 = 5 + 27 = 32.

Using the nth term formula, a + (n-1)d = 7 + 5*(-2) = 7 - 10 = -3.

Using the formula S_n = n/2 * (2a + (n-1)d), we have 100 = 10/2 * (2a + 9*2). Solving gives a = 10.

Using the formula S_n = n/2 * (2a + (n-1)d), we can substitute n = 10 and d = 5 to find a = 20.

Using the sum formula S_n = n/2 * (2a + (n-1)d), we have 50 = 5/2 * (10 + 4d). Solving gives d = 7.

Using the sum formula S_n = n/2 * (2a + (n-1)d), we have 50 = 5/2 * (2a + 8). Solving gives a = 10.

The sum of the first n terms is given by S_n = n/2 * (2a + (n-1)d). Here, 50 = 5/2 * (2a + 8). Solving gives a = 10.

Using the formula for the nth term, the 15th term = 7 + (15-1) * 2 = 7 + 28 = 35.

The nth term is given by a + (n-1)d. Here, a = 10, d = 5, and n = 4. So, the 4th term = 10 + (4-1) * 5 = 10 + 15 = 25.

The sum of the first n terms S_n = n/2 * (2a + (n-1)d). Here, S_15 = 15/2 * (2*10 + 14*2) = 15/2 * (20 + 28) = 15/2 * 48 = 360.

The sum of the first n terms S_n = n/2 * (2a + (n-1)d). Here, S_15 = 15/2 * (2*2 + 14*4) = 15/2 * (4 + 56) = 15/2 * 60 = 450.

An arithmetic progression has a constant difference between consecutive terms. The sequence 1, 3, 5, 7 has a common difference of 2.

In an arithmetic progression, the difference between any two consecutive terms is indeed constant, which defines the sequence.

In an arithmetic progression, the difference between consecutive terms is constant, which is the defining property of an AP.