The problem given involves three consecutive terms of an arithmetic progression (AP):

$-2x - 1, \quad 3x + 2, \quad \text{and} \quad 6x + 3$

a) To find the common difference:

For an arithmetic progression, the difference between any two consecutive terms is the same, meaning:

$(3x + 2) - (-2x - 1) = (6x + 3) - (3x + 2)$

Let’s simplify both sides:

1. $(3x + 2) - (-2x - 1) = 3x + 2 + 2x + 1 = 5x + 3$
2. $(6x + 3) - (3x + 2) = 6x + 3 - 3x - 2 = 3x + 1$

Now, set the two expressions equal to each other since the differences must be the same:

$5x + 3 = 3x + 1$

Solve for $x$:

$5x - 3x = 1 - 3$ $2x = -2$ $x = -1$

b) Find the first term if $3x + 2$ is the third term:

Given that $3x + 2$ is the third term, we substitute $x = -1$ into the expression $3x + 2$:

$3(-1) + 2 = -3 + 2 = -1$

Now that we know the third term is $-1$, we can find the first term using the formula for the $n$-th term of an arithmetic progression:

The third term can be written as $a + 2d = -1$, where $a$ is the first term and $d$ is the common difference.

We already calculated the common difference from part (a):

$d = 3x + 1 = 3(-1) + 1 = -3 + 1 = -2$

Substitute this into the equation:

$a + 2(-2) = -1$ $a - 4 = -1$ $a = 3$

Thus, the first term is $a = 3$.

Final answers:

a) The common difference is $-2$.

b) The first term is $3$.

Do you want further details on this solution? Here are some related questions for more practice:

1. How do you verify if three terms form an arithmetic progression?
2. What is the formula for the nth term of an arithmetic progression?
3. Can the common difference in an arithmetic progression be positive?
4. What happens to an arithmetic progression if the common difference is 0?
5. How would you calculate the sum of the first $n$ terms of an arithmetic progression?

Tip: In an arithmetic progression, the differences between consecutive terms are always equal, allowing you to set up equations and solve for unknowns efficiently.