4 Calculate: $$ \frac{4+6+8 \ldots+806}{5+10+15+\ldots+625} $$ 5 If the 7 th term of an arithmetic series is -3 and the sum of the first 16 terms is 24 , determine the first term. 6 The first term of an arithmetic series is 5 and the last term is 61 . The sum of all the terms is 957 . Calculate the common difference.

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Let's solve **Problem 6** step by step.

**Problem 6:**
*The first term of an arithmetic series is 5 and the last term is 61. The sum of all the terms is 957. Calculate the common difference.*

**Given:**
- First term ($$a$$) = 5
- Last term ($$a_n$$) = 61
- Sum of the series ($$S_n$$) = 957

**Objective:**
Find the common difference ($$d$$).

**Step 1: Find the Number of Terms ($$n$$)**

The $$n$$-th term of an arithmetic series is given by:
$$
a_n = a + (n - 1)d
$$
Plugging in the known values:
$$
61 = 5 + (n - 1)d \quad \Rightarrow \quad (n - 1)d = 56 \quad \Rightarrow \quad n - 1 = \frac{56}{d} \quad \Rightarrow \quad n = \frac{56}{d} + 1
$$

**Step 2: Use the Sum Formula**

The sum of the first $$n$$ terms of an arithmetic series is:
$$
S_n = \frac{n}{2} (a + a_n)
$$
Substituting the known values:
$$
957 = \frac{n}{2} (5 + 61) \quad \Rightarrow \quad 957 = \frac{n}{2} \times 66 \quad \Rightarrow \quad 957 = 33n \quad \Rightarrow \quad n = \frac{957}{33} = 29
$$

**Step 3: Find the Common Difference ($$d$$)**

Now that we know $$n = 29$$, substitute back into the equation from Step 1:
$$
29 = \frac{56}{d} + 1 \quad \Rightarrow \quad 28 = \frac{56}{d} \quad \Rightarrow \quad d = \frac{56}{28} = 2
$$

**Answer:**
The common difference is **2**.

**Summary of Solutions:**

1. **Problem 4:**
   $$
   \frac{4 + 6 + 8 + \ldots + 806}{5 + 10 + 15 + \ldots + 625} = \frac{3618}{875}
   $$

2. **Problem 5:**
   The first term is **-21**.

3. **Problem 6:**
   The common difference is **2**.
**Problem 6 Solution:** The common difference is **2**.

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