Verified answer

Answer:

Let's denote the first term of the parallel series as "a" and the common difference between consecutive terms as "d."

We know that the sum of an arithmetic series can be calculated using the formula:

Sum = (n/2)[2a + (n-1)d]

In this case, we have 21 positions, so n = 21.

Now, we are given two pieces of information:

1. The sum of the 10th, 11th, and 12th terms is 129. Using this information:

Sum(10th, 11th, 12th terms) = (3/2)[2a + (3-1)d] = 129

2. The sum of the last three terms is 237. Using this information:

Sum(last 3 terms) = (3/2)[2a + (3-1)d] = 237

Now, we can set up a system of equations with these two equations:

1. (3/2)[2a + 2d] = 129

2. (3/2)[2a + 2(d*(21-3))] = 237

Simplify these equations:

1. 3(a + d) = 129

2. 3(a + 36d) = 237

Now, solve this system of equations. First, solve equation 1 for "a":

a + d = 43

Now, substitute this value for "a" into equation 2:

3(43 + 36d) = 237

Simplify:

43 + 36d = 79

Subtract 43 from both sides:

36d = 79 - 43

36d = 36

Divide by 36:

d = 1

Now that we have found "d," we can find "a" using the equation a + d = 43:

a + 1 = 43

Subtract 1 from both sides:

a = 42

So, the first term "a" is 42, and the common difference "d" is 1.

Now, we can list the terms of the parallel series:

42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62

These are the 21 terms of the parallel series.