Let's denote the following variables:

• $a$ as the first term of the arithmetic sequence.
• $d$ as the common difference.
• $n$ as the number of terms in the sequence.

The formula for the $n$-th term of an arithmetic sequence is:

$T_n = a + (n-1)d$

We know that the eleventh term ($T_{11}$) is 30. Using the formula for the $n$-th term:

$T_{11} = a + 10d = 30 \quad \text{(1)}$

Next, the formula for the sum of the first $n$ terms of an arithmetic sequence is:

$S_n = \frac{n}{2} \times (2a + (n-1)d)$

We know the sum of the first eleven terms ($S_{11}$) is 55. Substituting into the formula:

$S_{11} = \frac{11}{2} \times (2a + 10d) = 55 \quad \text{(2)}$

Now, let's solve these two equations:

1. From equation (1): $a + 10d = 30$.
2. From equation (2): $\frac{11}{2} \times (2a + 10d) = 55$.

We can solve this system of equations to find the common difference $d$.

Solving the system

First, simplify equation (2):

$\frac{11}{2} \times (2a + 10d) = 55$

Multiply both sides by 2 to get rid of the fraction:

$11(2a + 10d) = 110$

Now divide both sides by 11:

$2a + 10d = 10 \quad \text{(3)}$

Now we have the system:

1. $a + 10d = 30$
2. $2a + 10d = 10$

Subtract the first equation from the second:

$(2a + 10d) - (a + 10d) = 10 - 30$

This simplifies to:

$a = -20$

Now substitute $a = -20$ into equation (1):

$-20 + 10d = 30$

Solving for $d$:

$10d = 50$ $d = 5$

Final Answer:

The common difference $d$ is $\boxed{5}$.

Would you like further details or explanations?

Here are five related questions you can explore:

1. How do you calculate the sum of the first $n$ terms in an arithmetic sequence?
2. What happens if the common difference in an arithmetic sequence is negative?
3. How do you find the first term of an arithmetic sequence given other terms?
4. What is the formula for the $n$-th term in an arithmetic sequence?
5. Can the sum of an arithmetic sequence be negative if all terms are positive?

Tip: Always double-check your work by substituting your values back into the original equations to ensure consistency!