Let’s solve the problems step by step.

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: $\boxed{1215}$

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: