Comprehensive worksheet designed to test your understanding of arithmetic and geometric sequences through varied problems.

Let's solve each problem step by step:

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Problem 1: What is the common difference of the arithmetic series 4, 7, 10, 13, …?

#### Solution:
The common difference \( d \) of an arithmetic series is the difference between any two consecutive terms. For the given series:
\[ d = 7 - 4 = 3 \]
\[ d = 10 - 7 = 3 \]
\[ d = 13 - 10 = 3 \]

Thus, the common difference is:
\[ \boxed{3} \]

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Problem 2: Find the 11th term of the series 2, 7, 12, 17, …

#### Solution:
The general formula for the \( n \)-th term of an arithmetic series is:
\[ a_n = a + (n-1)d \]
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

For the given series:
- First term \( a = 2 \),
- Common difference \( d = 7 - 2 = 5 \).

To find the 11th term (\( n = 11 \)):
\[ a_{11} = 2 + (11-1) \cdot 5 \]
\[ a_{11} = 2 + 10 \cdot 5 \]
\[ a_{11} = 2 + 50 \]
\[ a_{11} = 52 \]

Thus, the 11th term is:
\[ \boxed{52} \]

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Problem 3: What is the difference between the 4th and 7th term of the series 5, 10, 15, 20, …?

#### Solution:
The general formula for the \( n \)-th term of an arithmetic series is:
\[ a_n = a + (n-1)d \]

For the given series:
- First term \( a = 5 \),
- Common difference \( d = 10 - 5 = 5 \).

First, find the 4th term (\( n = 4 \)):
\[ a_4 = 5 + (4-1) \cdot 5 \]
\[ a_4 = 5 + 3 \cdot 5 \]
\[ a_4 = 5 + 15 \]
\[ a_4 = 20 \]

Next, find the 7th term (\( n = 7 \)):
\[ a_7 = 5 + (7-1) \cdot 5 \]
\[ a_7 = 5 + 6 \cdot 5 \]
\[ a_7 = 5 + 30 \]
\[ a_7 = 35 \]

Now, find the difference between the 7th and 4th terms:
\[ \text{Difference} = a_7 - a_4 \]
\[ \text{Difference} = 35 - 20 \]
\[ \text{Difference} = 15 \]

Thus, the difference is:
\[ \boxed{15} \]

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Problem 4: The first term of the series is 8 and the common difference is 7. Find the first five terms of the arithmetic series.

#### Solution:
The general formula for the \( n \)-th term of an arithmetic series is:
\[ a_n = a + (n-1)d \]

Given:
- First term \( a = 8 \),
- Common difference \( d = 7 \).

Find the first five terms:
1. \( a_1 = 8 \)
2. \( a_2 = 8 + (2-1) \cdot 7 = 8 + 7 = 15 \)
3. \( a_3 = 8 + (3-1) \cdot 7 = 8 + 2 \cdot 7 = 8 + 14 = 22 \)
4. \( a_4 = 8 + (4-1) \cdot 7 = 8 + 3 \cdot 7 = 8 + 21 = 29 \)
5. \( a_5 = 8 + (5-1) \cdot 7 = 8 + 4 \cdot 7 = 8 + 28 = 36 \)

Thus, the first five terms are:
\[ \boxed{8, 15, 22, 29, 36} \]

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Problem 5: Find the next three terms of the series 7, 17, 27, 37, …

#### Solution:
The given series is an arithmetic series with:
- First term \( a = 7 \),
- Common difference \( d = 17 - 7 = 10 \).

The general formula for the \( n \)-th term is:
\[ a_n = a + (n-1)d \]

Find the next three terms:
1. Sixth term (\( n = 6 \)):
\[ a_6 = 7 + (6-1) \cdot 10 = 7 + 5 \cdot 10 = 7 + 50 = 57 \]

2. Seventh term (\( n = 7 \)):
\[ a_7 = 7 + (7-1) \cdot 10 = 7 + 6 \cdot 10 = 7 + 60 = 67 \]

3. Eighth term (\( n = 8 \)):
\[ a_8 = 7 + (8-1) \cdot 10 = 7 + 7 \cdot 10 = 7 + 70 = 77 \]

Thus, the next three terms are:
\[ \boxed{47, 57, 67} \]

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Problem 6: What is the sum of the first 10 terms of the series 7, 11, 15, 19, …?

#### Solution:
The sum of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \]

For the given series:
- First term \( a = 7 \),
- Common difference \( d = 11 - 7 = 4 \),
- Number of terms \( n = 10 \).

Substitute into the formula:
\[ S_{10} = \frac{10}{2} \left( 2 \cdot 7 + (10-1) \cdot 4 \right) \]
\[ S_{10} = 5 \left( 14 + 9 \cdot 4 \right) \]
\[ S_{10} = 5 \left( 14 + 36 \right) \]
\[ S_{10} = 5 \cdot 50 \]
\[ S_{10} = 250 \]

Thus, the sum of the first 10 terms is:
\[ \boxed{250} \]

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Problem 7: In an arithmetic series, the 6th term is 32 and the common difference is 5. Find the first three terms of the series.

#### Solution:
The general formula for the \( n \)-th term of an arithmetic series is:
\[ a_n = a + (n-1)d \]

Given:
- 6th term \( a_6 = 32 \),
- Common difference \( d = 5 \).

Using the formula for the 6th term:
\[ a_6 = a + (6-1)d \]
\[ 32 = a + 5 \cdot 5 \]
\[ 32 = a + 25 \]
\[ a = 32 - 25 \]
\[ a = 7 \]

Now, find the first three terms:
1. First term (\( n = 1 \)):
\[ a_1 = a = 7 \]

2. Second term (\( n = 2 \)):
\[ a_2 = a + (2-1)d = 7 + 1 \cdot 5 = 7 + 5 = 12 \]

3. Third term (\( n = 3 \)):
\[ a_3 = a + (3-1)d = 7 + 2 \cdot 5 = 7 + 10 = 17 \]

Thus, the first three terms are:
\[ \boxed{7, 12, 17} \]

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Problem 8: The sum of the \( n \) terms of the series 61, 65, 69, 73, … is equal to 600. Find the value of \( n \).

#### Solution:
The sum of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \]

For the given series:
- First term \( a = 61 \),
- Common difference \( d = 65 - 61 = 4 \),
- Sum of \( n \) terms \( S_n = 600 \).

Substitute into the formula:
\[ 600 = \frac{n}{2} \left( 2 \cdot 61 + (n-1) \cdot 4 \right) \]
\[ 600 = \frac{n}{2} \left( 122 + 4(n-1) \right) \]
\[ 600 = \frac{n}{2} \left( 122 + 4n - 4 \right) \]
\[ 600 = \frac{n}{2} \left( 4n + 118 \right) \]
\[ 600 = \frac{n(4n + 118)}{2} \]
\[ 1200 = n(4n + 118) \]
\[ 4n^2 + 118n - 1200 = 0 \]

Divide the entire equation by 2:
\[ 2n^2 + 59n - 600 = 0 \]

Solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = 59 \), and \( c = -600 \):
\[ n = \frac{-59 \pm \sqrt{59^2 - 4 \cdot 2 \cdot (-600)}}{2 \cdot 2} \]
\[ n = \frac{-59 \pm \sqrt{3481 + 4800}}{4} \]
\[ n = \frac{-59 \pm \sqrt{8281}}{4} \]
\[ n = \frac{-59 \pm 91}{4} \]

This gives two solutions:
\[ n = \frac{-59 + 91}{4} = \frac{32}{4} = 8 \]
\[ n = \frac{-59 - 91}{4} = \frac{-150}{4} = -37.5 \] (not valid since \( n \) must be positive)

Thus, \( n = 8 \).

The value of \( n \) is:
\[ \boxed{8} \]

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Problem 9: The 20th and 30th terms of the arithmetic series are 120 and 180, respectively. Find the first two terms of this series.

#### Solution:
The general formula for the \( n \)-th term of an arithmetic series is:
\[ a_n = a + (n-1)d \]

Given:
- 20th term \( a_{20} = 120 \),
- 30th term \( a_{30} = 180 \).

Using the formula for the 20th term:
\[ a_{20} = a + (20-1)d \]
\[ 120 = a + 19d \quad \text{(Equation 1)} \]

Using the formula for the 30th term:
\[ a_{30} = a + (30-1)d \]
\[ 180 = a + 29d \quad \text{(Equation 2)} \]

Subtract Equation 1 from Equation 2:
\[ 180 - 120 = (a + 29d) - (a + 19d) \]
\[ 60 = 10d \]
\[ d = 6 \]

Substitute \( d = 6 \) back into Equation 1:
\[ 120 = a + 19 \cdot 6 \]
\[ 120 = a + 114 \]
\[ a = 120 - 114 \]
\[ a = 6 \]

Now, find the second term (\( n = 2 \)):
\[ a_2 = a + (2-1)d = 6 + 1 \cdot 6 = 6 + 6 = 12 \]

Thus, the first two terms are:
\[ \boxed{6, 12} \]

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Problem 10: The common ratio of the geometric series 7, 35, 175, … is ______

#### Solution:
The common ratio \( r \) of a geometric series is the ratio of any term to its preceding term. For the given series:
\[ r = \frac{35}{7} = 5 \]
\[ r = \frac{175}{35} = 5 \]

Thus, the common ratio is:
\[ \boxed{5} \]

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Final Answers:

1. \( \boxed{3} \)
2. \( \boxed{52} \)
3. \( \boxed{15} \)
4. \( \boxed{8, 15, 22, 29, 36} \)
5. \( \boxed{47, 57, 67} \)
6. \( \boxed{250} \)
7. \( \boxed{7, 12, 17} \)
8. \( \boxed{8} \)
9. \( \boxed{6, 12} \)
10. \( \boxed{5} \)