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Let’s solve the problems step by step.

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### **Problem 6: Arithmetic Sum**

The given sequence is:
\[ 8 + 15 + 22 + 29 + \dots + 106 + 113 + 120 + 127 \]

This is an **arithmetic sequence** with:
- First term \(a = 8\),
- Common difference \(d = 7\),
- Last term \(l = 127\).

To solve for the sum, we need the number of terms \(n\) in the sequence:
\[
l = a + (n - 1)d
\]
Substitute \(l = 127\), \(a = 8\), and \(d = 7\):
\[
127 = 8 + (n - 1) \cdot 7
\]
\[
127 - 8 = 7(n - 1)
\]
\[
119 = 7(n - 1)
\]
\[
n - 1 = 17 \quad \Rightarrow \quad n = 18
\]

Now, use the sum formula for an arithmetic sequence:
\[
S_n = \frac{n}{2}(a + l)
\]
Substitute \(n = 18\), \(a = 8\), and \(l = 127\):
\[
S_{18} = \frac{18}{2}(8 + 127)
\]
\[
S_{18} = 9 \cdot 135 = 1215
\]

**Answer for Problem 6:**
\[
\boxed{1215}
\]

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### **Problem 7: Sum of Fractions**

The given series is:
\[
\frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \frac{1}{9 \cdot 11} + \dots + \frac{1}{31 \cdot 33} + \frac{1}{33 \cdot 35}
\]

This series is a **telescoping series**, where:
\[
\frac{1}{n(n + 2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n + 2} \right)
\]

Rewrite each term:
\[
\frac{1}{5 \cdot 7} = \frac{1}{2} \left( \frac{1}{5} - \frac{1}{7} \right), \quad
\frac{1}{7 \cdot 9} = \frac{1}{2} \left( \frac{1}{7} - \frac{1}{9} \right), \quad \dots
\]

The general term is:
\[
\frac{1}{n(n + 2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n + 2} \right)
\]

The series becomes:
\[
\frac{1}{2} \left( \frac{1}{5} - \frac{1}{7} \right) + \frac{1}{2} \left( \frac{1}{7} - \frac{1}{9} \right) + \dots + \frac{1}{2} \left( \frac{1}{33} - \frac{1}{35} \right)
\]

Notice that this is a telescoping sum:
\[
\frac{1}{2} \left( \frac{1}{5} - \frac{1}{35} \right)
\]

Now calculate:
\[
\frac{1}{2} \left( \frac{1}{5} - \frac{1}{35} \right) = \frac{1}{2} \left( \frac{7 - 1}{35} \right) = \frac{1}{2} \cdot \frac{6}{35} = \frac{3}{35}
\]

**Answer for Problem 7:**
\[
\boxed{\frac{3}{35}}
\]

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Do you have any questions or would you like further clarification?  

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### **Related Questions**
1. What is the formula for the \(n\)-th term of an arithmetic sequence?
2. How do you derive the formula for the sum of an arithmetic sequence?
3. What is the principle of telescoping series in fractions?
4. Can you generalize the sum of fractions of the form \( \frac{1}{n(n+k)} \)?
5. How do you handle arithmetic progressions with fractional or negative terms?

### **Tip:**  
For telescoping series, always simplify the general term and look for patterns that cancel across the sequence.

6. Find the value of 8 + 15 + 22 + 29 + ... + 106 + 113 + 120 + 127. 7. Find the value of 1/(5×7) + 1/(7×9) + 1/(9×11) + ... + 1/(31×33) + 1/(33×35).

Learn how to calculate the sum of an arithmetic sequence and evaluate a telescoping series. This step-by-step solution explains the use of key formulas, including the arithmetic sequence sum formula and telescoping series simplification. Suitable for Grades 9-11 students.

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