1-4 Additional Practice Arithmetic Sequences And Series Answer Key

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Mar 15, 2025 · 5 min read

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1-4 Additional Practice Arithmetic Sequences and Series: Answer Key & Explanations
This comprehensive guide provides detailed solutions and explanations for four additional practice problems involving arithmetic sequences and series. Understanding arithmetic sequences and series is crucial for success in algebra and beyond, forming the foundation for more advanced mathematical concepts. This guide aims to solidify your understanding through worked examples and insightful explanations, helping you confidently tackle similar problems.
Problem 1: Finding the nth Term and the Sum
Question: An arithmetic sequence begins with the terms 5, 11, 17, ... Find:
a) The 20th term of the sequence (a<sub>20</sub>).
b) The sum of the first 20 terms (S<sub>20</sub>).
Solution:
a) Finding the 20th term (a<sub>20</sub>):
First, we identify the common difference (d) between consecutive terms: d = 11 - 5 = 6.
The formula for the nth term of an arithmetic sequence is: a<sub>n</sub> = a<sub>1</sub> + (n-1)d, where a<sub>1</sub> is the first term, n is the term number, and d is the common difference.
Plugging in the values for our problem (a<sub>1</sub> = 5, n = 20, d = 6):
a<sub>20</sub> = 5 + (20-1)6 = 5 + 19 * 6 = 5 + 114 = 119
Therefore, the 20th term of the sequence is $\boxed{119}$.
b) Finding the sum of the first 20 terms (S<sub>20</sub>):
The formula for the sum of the first n terms of an arithmetic series is: S<sub>n</sub> = n/2 [2a<sub>1</sub> + (n-1)d] or S<sub>n</sub> = n/2 (a<sub>1</sub> + a<sub>n</sub>).
Using the first formula:
S<sub>20</sub> = 20/2 [2(5) + (20-1)6] = 10 [10 + 114] = 10 * 124 = 1240
Using the second formula (since we already know a<sub>20</sub> = 119):
S<sub>20</sub> = 20/2 (5 + 119) = 10 * 124 = 1240
Therefore, the sum of the first 20 terms is $\boxed{1240}$.
Problem 2: Finding the First Term and Common Difference
Question: The 7th term of an arithmetic sequence is 34 and the 12th term is 64. Find the first term (a<sub>1</sub>) and the common difference (d).
Solution:
We have two equations based on the formula a<sub>n</sub> = a<sub>1</sub> + (n-1)d:
- a<sub>7</sub> = a<sub>1</sub> + 6d = 34
- a<sub>12</sub> = a<sub>1</sub> + 11d = 64
We can solve this system of equations using subtraction. Subtract the first equation from the second:
(a<sub>1</sub> + 11d) - (a<sub>1</sub> + 6d) = 64 - 34
This simplifies to: 5d = 30, so d = 6.
Now, substitute d = 6 back into either of the original equations to solve for a<sub>1</sub>. Let's use the first equation:
a<sub>1</sub> + 6(6) = 34
a<sub>1</sub> + 36 = 34
a<sub>1</sub> = 34 - 36 = -2
Therefore, the first term is $\boxed{-2}$ and the common difference is $\boxed{6}$.
Problem 3: Working with a Given Sum
Question: The sum of the first 15 terms of an arithmetic series is 600, and the first term is 20. Find the common difference (d).
Solution:
We use the formula for the sum of an arithmetic series: S<sub>n</sub> = n/2 [2a<sub>1</sub> + (n-1)d].
We know: S<sub>15</sub> = 600, a<sub>1</sub> = 20, and n = 15. We need to find d.
600 = 15/2 [2(20) + (15-1)d]
600 = 7.5 [40 + 14d]
Multiply both sides by 2/15 to simplify:
80 = 40 + 14d
Subtract 40 from both sides:
40 = 14d
d = 40/14 = 20/7
Therefore, the common difference is $\boxed{20/7}$.
Problem 4: Applications of Arithmetic Sequences
Question: A construction worker is stacking bricks. On the first day, he stacks 50 bricks. Each day thereafter, he stacks 10 fewer bricks than the previous day. How many bricks will he stack in total after 8 days?
Solution:
This problem describes an arithmetic sequence where the first term (a<sub>1</sub>) is 50, and the common difference (d) is -10 (since he stacks 10 fewer bricks each day). We need to find the sum of the first 8 terms (S<sub>8</sub>).
Using the formula S<sub>n</sub> = n/2 [2a<sub>1</sub> + (n-1)d]:
S<sub>8</sub> = 8/2 [2(50) + (8-1)(-10)]
S<sub>8</sub> = 4 [100 + 7(-10)]
S<sub>8</sub> = 4 [100 - 70]
S<sub>8</sub> = 4 * 30 = 120
Therefore, the construction worker will stack a total of $\boxed{120}$ bricks after 8 days.
Further Practice and Key Concepts to Remember
These four problems illustrate the fundamental concepts of arithmetic sequences and series. To further enhance your understanding, consider these points:
-
Identifying Arithmetic Sequences: Always check if the difference between consecutive terms remains constant. This constant difference is the common difference (d).
-
Mastering the Formulas: Familiarize yourself with the formulas for the nth term (a<sub>n</sub> = a<sub>1</sub> + (n-1)d) and the sum of the first n terms (S<sub>n</sub> = n/2 [2a<sub>1</sub> + (n-1)d] or S<sub>n</sub> = n/2 (a<sub>1</sub> + a<sub>n</sub>)).
-
Solving Systems of Equations: As demonstrated in Problem 2, you'll often need to solve systems of equations to find missing variables. Practice different methods, such as substitution or elimination.
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Real-World Applications: Arithmetic sequences and series appear in various real-world scenarios, from stacking objects to calculating financial growth (with slight modifications for compound interest). Try applying these concepts to different contexts.
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Practice, Practice, Practice: Work through additional problems to reinforce your understanding and build confidence.
By understanding these key concepts and practicing regularly, you will master arithmetic sequences and series, opening the door to more advanced mathematical concepts. Remember to always break down complex problems into smaller, manageable steps, and don't hesitate to review the formulas and techniques as needed. Consistent practice is the key to success in mathematics.
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