Answer :
* Problem 17: Cannot determine the common difference with the given information.
* Problem 18: Substitute $a=3$, $d=2$, and $S_n = 120$ into the sum formula, solve the quadratic equation $n^2 + 2n - 120 = 0$, and find $n=10$.
* Problem 19: Use the sum formula with $a=5$, $d=5$, and $n=10$ to calculate the sum $S_{10} = 275$.
* The number of terms for Problem 18 is $\boxed{10}$ and the sum for Problem 19 is $\boxed{275}$.
### Explanation
1. Introduction
Let's break down each problem step by step to make sure we understand everything clearly.
2. Problem 17 Analysis
**Problem 17: Find the common difference of the AP with last term -94.**
Unfortunately, we can't find the common difference with just the last term. We need more information, like the first term or the number of terms, to solve this.
3. Problem 18 Solution
**Problem 18: How many terms of AP $3, 5, 7, 9, \ldots$ must be taken to get the sum 120?**
Here, we have an arithmetic progression (AP) with the first term $a = 3$ and a common difference $d = 2$. We want to find the number of terms, $n$, such that the sum of the first $n$ terms, $S_n$, is equal to 120.
The formula for the sum of the first $n$ terms of an AP is:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
Plugging in the values $a = 3$, $d = 2$, and $S_n = 120$, we get:
$$120 = \frac{n}{2}[2(3) + (n-1)2]$$
$$120 = \frac{n}{2}[6 + 2n - 2]$$
$$120 = \frac{n}{2}[4 + 2n]$$
$$240 = n[4 + 2n]$$
$$240 = 4n + 2n^2$$
Rearranging the equation, we get a quadratic equation:
$$2n^2 + 4n - 240 = 0$$
Dividing by 2:
$$n^2 + 2n - 120 = 0$$
Factoring the quadratic equation:
$$(n - 10)(n + 12) = 0$$
The possible values for $n$ are $n = 10$ or $n = -12$. Since the number of terms cannot be negative, we have $n = 10$.
4. Problem 19 Solution
**Problem 19: Find the sum of the first ten multiples of 5.**
The first ten multiples of 5 are: $5, 10, 15, 20, 25, 30, 35, 40, 45, 50$. This is an AP with first term $a = 5$, common difference $d = 5$, and number of terms $n = 10$.
The sum of the first $n$ terms of an AP is given by:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
Plugging in the values $a = 5$, $d = 5$, and $n = 10$, we get:
$$S_{10} = \frac{10}{2}[2(5) + (10-1)5]$$
$$S_{10} = 5[10 + 9(5)]$$
$$S_{10} = 5[10 + 45]$$
$$S_{10} = 5[55]$$
$$S_{10} = 275$$
5. Conclusion
**Final Answers:**
* Problem 17: Cannot be determined 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
Understanding arithmetic progressions can be incredibly useful in everyday life. For instance, if you're saving money each month with a fixed increase, you can use the sum of an AP to calculate your total savings after a certain period. Similarly, if you're planning a workout routine with increasing intensity, APs can help you track your progress and plan your future workouts effectively. These mathematical tools provide a structured way to analyze and predict outcomes in various real-world scenarios.
* Problem 18: Substitute $a=3$, $d=2$, and $S_n = 120$ into the sum formula, solve the quadratic equation $n^2 + 2n - 120 = 0$, and find $n=10$.
* Problem 19: Use the sum formula with $a=5$, $d=5$, and $n=10$ to calculate the sum $S_{10} = 275$.
* The number of terms for Problem 18 is $\boxed{10}$ and the sum for Problem 19 is $\boxed{275}$.
### Explanation
1. Introduction
Let's break down each problem step by step to make sure we understand everything clearly.
2. Problem 17 Analysis
**Problem 17: Find the common difference of the AP with last term -94.**
Unfortunately, we can't find the common difference with just the last term. We need more information, like the first term or the number of terms, to solve this.
3. Problem 18 Solution
**Problem 18: How many terms of AP $3, 5, 7, 9, \ldots$ must be taken to get the sum 120?**
Here, we have an arithmetic progression (AP) with the first term $a = 3$ and a common difference $d = 2$. We want to find the number of terms, $n$, such that the sum of the first $n$ terms, $S_n$, is equal to 120.
The formula for the sum of the first $n$ terms of an AP is:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
Plugging in the values $a = 3$, $d = 2$, and $S_n = 120$, we get:
$$120 = \frac{n}{2}[2(3) + (n-1)2]$$
$$120 = \frac{n}{2}[6 + 2n - 2]$$
$$120 = \frac{n}{2}[4 + 2n]$$
$$240 = n[4 + 2n]$$
$$240 = 4n + 2n^2$$
Rearranging the equation, we get a quadratic equation:
$$2n^2 + 4n - 240 = 0$$
Dividing by 2:
$$n^2 + 2n - 120 = 0$$
Factoring the quadratic equation:
$$(n - 10)(n + 12) = 0$$
The possible values for $n$ are $n = 10$ or $n = -12$. Since the number of terms cannot be negative, we have $n = 10$.
4. Problem 19 Solution
**Problem 19: Find the sum of the first ten multiples of 5.**
The first ten multiples of 5 are: $5, 10, 15, 20, 25, 30, 35, 40, 45, 50$. This is an AP with first term $a = 5$, common difference $d = 5$, and number of terms $n = 10$.
The sum of the first $n$ terms of an AP is given by:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
Plugging in the values $a = 5$, $d = 5$, and $n = 10$, we get:
$$S_{10} = \frac{10}{2}[2(5) + (10-1)5]$$
$$S_{10} = 5[10 + 9(5)]$$
$$S_{10} = 5[10 + 45]$$
$$S_{10} = 5[55]$$
$$S_{10} = 275$$
5. Conclusion
**Final Answers:**
* Problem 17: Cannot be determined 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
Understanding arithmetic progressions can be incredibly useful in everyday life. For instance, if you're saving money each month with a fixed increase, you can use the sum of an AP to calculate your total savings after a certain period. Similarly, if you're planning a workout routine with increasing intensity, APs can help you track your progress and plan your future workouts effectively. These mathematical tools provide a structured way to analyze and predict outcomes in various real-world scenarios.
