Answer :
* Problem 17 cannot be solved because the common difference cannot be determined with only one term.
* Problem 18 requires solving a quadratic equation to find the number of terms: $n^2 + 2n - 120 = 0$, which factors to $(n-10)(n+12) = 0$. Thus, $n = 10$.
* Problem 19 involves finding the sum of the first 10 multiples of 5 using the arithmetic series formula, resulting in $S_{10} = 275$.
* The number of terms is $\boxed{10}$ and the sum of the first ten multiples of 5 is $\boxed{275}$.
### Explanation
1. Solving the Problems
**Problem 17: Finding the Common Difference**
In an arithmetic progression (AP), the common difference is the constant difference between consecutive terms. To find the common difference, we need at least two terms of the AP. In this problem, we are given only the first term, which is -94. Without any other terms or information, it is impossible to determine the common difference. Therefore, we cannot solve this problem with the given information.
**Problem 18: Finding the Number of Terms**
We are given an arithmetic progression (AP) $3, 5, 7, 9, \ldots$ and we need to find how many terms must be taken to get a sum of 120. The first term, $a$, is 3, and the common difference, $d$, is $5 - 3 = 2$. The sum of the first $n$ terms of an AP is given by the formula:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
We are given that $S_n = 120$. Substituting the values of $a$ and $d$ into the formula, we get:
$$120 = \frac{n}{2}[2(3) + (n-1)2]$$
Simplifying the equation:
$$120 = \frac{n}{2}[6 + 2n - 2]$$
$$120 = \frac{n}{2}[4 + 2n]$$
$$120 = n(2 + n)$$
$$n^2 + 2n - 120 = 0$$
Now, we solve the quadratic equation $n^2 + 2n - 120 = 0$. We can factor this equation as $(n - 10)(n + 12) = 0$. The solutions are $n = 10$ and $n = -12$. Since $n$ must be a positive integer, we have $n = 10$.
Therefore, 10 terms of the AP must be taken to get a sum of 120.
**Problem 19: Finding the Sum of the First Ten Multiples of 5**
We need to find the sum of the first ten multiples of 5. The multiples of 5 form an arithmetic progression: $5, 10, 15, 20, \ldots$. The first term, $a$, is 5, the common difference, $d$, is 5, and the number of terms, $n$, is 10. The sum of the first $n$ terms of an AP is given by the formula:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
Substituting the values of $a$, $d$, and $n$ into the formula, we get:
$$S_{10} = \frac{10}{2}[2(5) + (10-1)5]$$
$$S_{10} = 5[10 + 45]$$
$$S_{10} = 5[55]$$
$$S_{10} = 275$$
Therefore, the sum of the first ten multiples of 5 is 275.
2. Stating the Answers
**Final Answers:**
Problem 17: Cannot be solved with the given information.
Problem 18: 10 terms must be taken to get a sum of 120.
Problem 19: The sum of the first ten multiples of 5 is 275.
### Examples
Arithmetic progressions are useful in many real-life situations, such as calculating simple interest, predicting salary increases, and determining the number of seats in an auditorium. Understanding arithmetic progressions helps in making informed decisions in financial planning and resource allocation. For example, if you deposit a fixed amount of money into a savings account each month, the total amount saved over time forms an arithmetic progression.
* Problem 18 requires solving a quadratic equation to find the number of terms: $n^2 + 2n - 120 = 0$, which factors to $(n-10)(n+12) = 0$. Thus, $n = 10$.
* Problem 19 involves finding the sum of the first 10 multiples of 5 using the arithmetic series formula, resulting in $S_{10} = 275$.
* The number of terms is $\boxed{10}$ and the sum of the first ten multiples of 5 is $\boxed{275}$.
### Explanation
1. Solving the Problems
**Problem 17: Finding the Common Difference**
In an arithmetic progression (AP), the common difference is the constant difference between consecutive terms. To find the common difference, we need at least two terms of the AP. In this problem, we are given only the first term, which is -94. Without any other terms or information, it is impossible to determine the common difference. Therefore, we cannot solve this problem with the given information.
**Problem 18: Finding the Number of Terms**
We are given an arithmetic progression (AP) $3, 5, 7, 9, \ldots$ and we need to find how many terms must be taken to get a sum of 120. The first term, $a$, is 3, and the common difference, $d$, is $5 - 3 = 2$. The sum of the first $n$ terms of an AP is given by the formula:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
We are given that $S_n = 120$. Substituting the values of $a$ and $d$ into the formula, we get:
$$120 = \frac{n}{2}[2(3) + (n-1)2]$$
Simplifying the equation:
$$120 = \frac{n}{2}[6 + 2n - 2]$$
$$120 = \frac{n}{2}[4 + 2n]$$
$$120 = n(2 + n)$$
$$n^2 + 2n - 120 = 0$$
Now, we solve the quadratic equation $n^2 + 2n - 120 = 0$. We can factor this equation as $(n - 10)(n + 12) = 0$. The solutions are $n = 10$ and $n = -12$. Since $n$ must be a positive integer, we have $n = 10$.
Therefore, 10 terms of the AP must be taken to get a sum of 120.
**Problem 19: Finding the Sum of the First Ten Multiples of 5**
We need to find the sum of the first ten multiples of 5. The multiples of 5 form an arithmetic progression: $5, 10, 15, 20, \ldots$. The first term, $a$, is 5, the common difference, $d$, is 5, and the number of terms, $n$, is 10. The sum of the first $n$ terms of an AP is given by the formula:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
Substituting the values of $a$, $d$, and $n$ into the formula, we get:
$$S_{10} = \frac{10}{2}[2(5) + (10-1)5]$$
$$S_{10} = 5[10 + 45]$$
$$S_{10} = 5[55]$$
$$S_{10} = 275$$
Therefore, the sum of the first ten multiples of 5 is 275.
2. Stating the Answers
**Final Answers:**
Problem 17: Cannot be solved with the given information.
Problem 18: 10 terms must be taken to get a sum of 120.
Problem 19: The sum of the first ten multiples of 5 is 275.
### Examples
Arithmetic progressions are useful in many real-life situations, such as calculating simple interest, predicting salary increases, and determining the number of seats in an auditorium. Understanding arithmetic progressions helps in making informed decisions in financial planning and resource allocation. For example, if you deposit a fixed amount of money into a savings account each month, the total amount saved over time forms an arithmetic progression.
