1. **State the problem:** Find the sum of the arithmetic sequence from the 30th term to the 75th term. 2. **Formula for sum of arithmetic sequence:** The sum of terms from $n_1$ to $n_2$ in an arithmetic sequence is given by $$S = \frac{(n_2 - n_1 + 1)}{2} \times (a_{n_1} + a_{n_2})$$ where $a_{n_1}$ and $a_{n_2}$ are the terms at positions $n_1$ and $n_2$ respectively. 3. **Important rules:** - The number of terms summed is $n_2 - n_1 + 1$. - You need to know the first term and common difference to find $a_{n_1}$ and $a_{n_2}$. 4. **Find $a_{30}$ and $a_{75}$:** If the first term is $a_1$ and common difference is $d$, then $$a_n = a_1 + (n-1)d$$ So, $$a_{30} = a_1 + 29d$$ $$a_{75} = a_1 + 74d$$ 5. **Sum from 30th to 75th term:** Number of terms: $$75 - 30 + 1 = 46$$ Sum: $$S = \frac{46}{2} \times (a_1 + 29d + a_1 + 74d) = 23 \times (2a_1 + 103d)$$ 6. **Answer:** The sum from the 30th to the 75th term of the arithmetic sequence is $$\boxed{23 \times (2a_1 + 103d)}$$ --- **Next question:** By which term will your geometric sequence have at least 200 shapes? 1. **State the problem:** Find the smallest term number $n$ such that the geometric sequence term $a_n \geq 200$. 2. **Formula for geometric sequence term:** $$a_n = a_1 \times r^{n-1}$$ where $a_1$ is the first term and $r$ is the common ratio. 3. **Set inequality:** $$a_1 \times r^{n-1} \geq 200$$ 4. **Solve for $n$:** Divide both sides by $a_1$: $$r^{n-1} \geq \frac{200}{a_1}$$ Take logarithm base $r$ (assuming $r>0$, $r \neq 1$): $$n-1 \geq \log_r \left(\frac{200}{a_1}\right)$$ So, $$n \geq 1 + \log_r \left(\frac{200}{a_1}\right)$$ 5. **Answer:** The term number $n$ at which the geometric sequence has at least 200 shapes is $$\boxed{n = \left\lceil 1 + \log_r \left(\frac{200}{a_1}\right) \right\rceil}$$ where $\lceil \cdot \rceil$ denotes the ceiling function to get the smallest integer $n$ satisfying the inequality.