Determine $$ n $$ in each of the arithmetic series: a) $$ 16+19+22+\ldots $$ (to $$ n $$ terms) $$ =295 $$ b) $$ 6+17+28+\ldots $$ (to $$ n $$ terms) $$ =273 $$

Question

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To determine the number of terms $$ n $$ in each arithmetic series, we'll use the formula for the sum of the first $$ n $$ terms of an arithmetic series:

$$
S_n = \frac{n}{2} \left( 2a_1 + (n-1)d \right)
$$

where:
- $$ S_n $$ is the sum of the series,
- $$ a_1 $$ is the first term,
- $$ d $$ is the common difference,
- $$ n $$ is the number of terms.

### Problem a)

**Given:**
- Series: $$ 16 + 19 + 22 + \ldots $$
- Sum $$ S_n = 295 $$

**Steps:**

1. **Identify $$ a_1 $$ and $$ d $$:**
   $$
   a_1 = 16, \quad d = 19 - 16 = 3
   $$

2. **Set up the sum formula:**
   $$
   \frac{n}{2} \left( 2 \times 16 + (n-1) \times 3 \right) = 295
   $$
   $$
   \frac{n}{2} \left( 32 + 3n - 3 \right) = 295
   $$
   $$
   \frac{n}{2} (29 + 3n) = 295
   $$

3. **Solve for $$ n $$:**
   $$
   n(29 + 3n) = 590
   $$
   $$
   3n^2 + 29n - 590 = 0
   $$

4. **Apply the quadratic formula:**
   $$
   n = \frac{-29 \pm \sqrt{29^2 - 4 \times 3 \times (-590)}}{2 \times 3}
   $$
   $$
   n = \frac{-29 \pm \sqrt{841 + 7080}}{6}
   $$
   $$
   n = \frac{-29 \pm 89}{6}
   $$

5. **Determine the valid solution:**
   $$
   n = \frac{60}{6} = 10 \quad (\text{since $$ n $$ must be positive})
   $$

**Answer:**
$$ \boxed{10} $$

### Problem b)

**Given:**
- Series: $$ 6 + 17 + 28 + \ldots $$
- Sum $$ S_n = 273 $$

**Steps:**

1. **Identify $$ a_1 $$ and $$ d $$:**
   $$
   a_1 = 6, \quad d = 17 - 6 = 11
   $$

2. **Set up the sum formula:**
   $$
   \frac{n}{2} \left( 2 \times 6 + (n-1) \times 11 \right) = 273
   $$
   $$
   \frac{n}{2} \left( 12 + 11n - 11 \right) = 273
   $$
   $$
   \frac{n}{2} (1 + 11n) = 273
   $$

3. **Solve for $$ n $$:**
   $$
   n(1 + 11n) = 546
   $$
   $$
   11n^2 + n - 546 = 0
   $$

4. **Apply the quadratic formula:**
   $$
   n = \frac{-1 \pm \sqrt{1 + 4 \times 11 \times 546}}{2 \times 11}
   $$
   $$
   n = \frac{-1 \pm \sqrt{1 + 24024}}{22}
   $$
   $$
   n = \frac{-1 \pm 155}{22}
   $$

5. **Determine the valid solution:**
   $$
   n = \frac{154}{22} = 7 \quad (\text{since $$ n $$ must be positive})
   $$

**Answer:**
$$ \boxed{7} $$

Answer:
Problem a Answer:
$$ n = 10 $$

Problem b Answer:
$$ n = 7 $$
Problem a: $$ n = 10 $$ Problem b: $$ n = 7 $$

Determine in each of the arithmetic series:
a) (to terms)
b) (to terms)

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a) For the series

b) For the series

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