The problem asks to evaluate the following arithmetic series:

$\sum_{n=1}^{10} \left( -2 + \frac{4}{3}n \right)$

This represents the sum of terms from $n = 1$ to $n = 10$ of the sequence defined by the expression $-2 + \frac{4}{3}n$.

Steps for solution:

1. Recognize the Arithmetic Series Format: The series is in the form: $a_n = -2 + \frac{4}{3}n$ where the first term ($a_1$) is when $n=1$ and the last term ($a_{10}$) is when $n=10$.

2. Find the First and Last Terms:

   • First term ($a_1$) when $n=1$: $a_1 = -2 + \frac{4}{3}(1) = -2 + \frac{4}{3} = -\frac{6}{3} + \frac{4}{3} = -\frac{2}{3}$
   • Last term ($a_{10}$) when $n=10$: $a_{10} = -2 + \frac{4}{3}(10) = -2 + \frac{40}{3} = -\frac{6}{3} + \frac{40}{3} = \frac{34}{3}$

3. Use the Arithmetic Series Sum Formula: The sum $S_n$ of an arithmetic series is given by: $S_n = \frac{n}{2} \left(a_1 + a_n\right)$ where $n$ is the number of terms. In this case, $n = 10$, $a_1 = -\frac{2}{3}$, and $a_{10} = \frac{34}{3}$.

4. Calculate the Sum: $S_{10} = \frac{10}{2} \left(-\frac{2}{3} + \frac{34}{3}\right)$ Simplifying inside the parentheses: $-\frac{2}{3} + \frac{34}{3} = \frac{32}{3}$ Now, compute the sum: $S_{10} = 5 \times \frac{32}{3} = \frac{160}{3}$

Final Answer:

The sum of the series is $\frac{160}{3}$.

Would you like further details or any other questions answered?

Here are some related questions to explore:

1. How would the sum change if we increased the number of terms?
2. What is the general formula for the $n$-th term of an arithmetic sequence?
3. How do you derive the formula for the sum of an arithmetic series?
4. What would happen if the common difference in the series were negative?
5. How do you calculate the sum of an infinite arithmetic series?

Tip: When dealing with arithmetic series, always confirm the first term, common difference, and last term to simplify calculations.