Let's begin by addressing the third problem shown in the image:

Problem 3:

The partial sum in an arithmetic series with a first term $a_1 = 17$ and a common difference $d = 3$ is $S_n = 30705$. We need to determine how many terms are in the series, that is, find $n$.

Step-by-step solution:

The sum of an arithmetic series is given by the formula: $S_n = \frac{n}{2} \times (2a_1 + (n-1) d)$ Where:

• $S_n$ is the sum of the first $n$ terms,
• $a_1$ is the first term,
• $d$ is the common difference,
• $n$ is the number of terms.

We are given:

• $a_1 = 17$,
• $d = 3$,
• $S_n = 30705$.

Substitute these values into the sum formula: $30705 = \frac{n}{2} \times (2(17) + (n-1)(3))$ Simplify: $30705 = \frac{n}{2} \times (34 + 3n - 3)$ $30705 = \frac{n}{2} \times (31 + 3n)$ Multiply both sides by 2 to eliminate the fraction: $61410 = n(31 + 3n)$ Expand the equation: $61410 = 31n + 3n^2$ Rearrange into a standard quadratic form: $3n^2 + 31n - 61410 = 0$ Now we will solve this quadratic equation using the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Where $a = 3$, $b = 31$, and $c = -61410$.

Let's calculate the value of $n$.The two solutions for $n$ are $138$ and $-148.33$. Since the number of terms $n$ must be a positive integer, we take $n = 138$.

Answer:

There are 138 terms in the arithmetic series.

Would you like further details on this, or help with another question from the image?

Here are some related questions for practice:

1. How do you find the sum of the first $n$ terms in an arithmetic sequence?
2. What is the difference between an arithmetic and a geometric series?
3. How would the solution change if the common difference was negative?
4. How can you use the quadratic formula in different types of word problems?
5. How do you determine if a quadratic equation will have real solutions?

Tip: Always check if your solution makes sense, especially when dealing with physical quantities or counting terms in a sequence. For example, a negative number of terms wouldn't be valid!