SOLUTION: Arithmetic series worksheet - Studypool - Free Printable

Problem Analysis:

The worksheet focuses on arithmetic sequences and series. We will solve the problems step by step using the formulas provided:

1. Formula for the nth term of an arithmetic sequence:
\[
a_n = a_1 + (n-1)d
\]
where:
- \( a_n \) is the nth term,
- \( a_1 \) is the first term,
- \( d \) is the common difference,
- \( n \) is the position of the term.

2. Formula for the sum of the first \( n \) terms of an arithmetic sequence:
\[
S_n = \frac{n}{2} (a_1 + a_n)
\]
where:
- \( S_n \) is the sum of the first \( n \) terms,
- \( a_1 \) is the first term,
- \( a_n \) is the nth term,
- \( n \) is the number of terms.

---

Problem 1: Write down the stated term and the formula for the nth term of the following arithmetic sequences

#### (a) Sequence: \( 7, 11, 15, \ldots \) (7th term)

1. Identify \( a_1 \), \( d \), and \( n \):
- First term: \( a_1 = 7 \)
- Common difference: \( d = 11 - 7 = 4 \)
- Position: \( n = 7 \)

2. Find the 7th term (\( a_7 \)):
\[
a_n = a_1 + (n-1)d
\]
\[
a_7 = 7 + (7-1) \cdot 4 = 7 + 6 \cdot 4 = 7 + 24 = 31
\]

3. Formula for the nth term:
\[
a_n = 7 + (n-1) \cdot 4
\]

Answer for (a):
\[
\boxed{31, \, a_n = 7 + (n-1) \cdot 4}
\]

#### (b) Sequence: \( -7, -5, -3, \ldots \) (23rd term)

1. Identify \( a_1 \), \( d \), and \( n \):
- First term: \( a_1 = -7 \)
- Common difference: \( d = -5 - (-7) = 2 \)
- Position: \( n = 23 \)

2. Find the 23rd term (\( a_{23} \)):
\[
a_n = a_1 + (n-1)d
\]
\[
a_{23} = -7 + (23-1) \cdot 2 = -7 + 22 \cdot 2 = -7 + 44 = 37
\]

3. Formula for the nth term:
\[
a_n = -7 + (n-1) \cdot 2
\]

Answer for (b):
\[
\boxed{37, \, a_n = -7 + (n-1) \cdot 2}
\]

#### (c) Sequence: \( 18, 11, 4, \ldots \) (6th term)

1. Identify \( a_1 \), \( d \), and \( n \):
- First term: \( a_1 = 18 \)
- Common difference: \( d = 11 - 18 = -7 \)
- Position: \( n = 6 \)

2. Find the 6th term (\( a_6 \)):
\[
a_n = a_1 + (n-1)d
\]
\[
a_6 = 18 + (6-1) \cdot (-7) = 18 + 5 \cdot (-7) = 18 - 35 = -17
\]

3. Formula for the nth term:
\[
a_n = 18 + (n-1) \cdot (-7)
\]

Answer for (c):
\[
\boxed{-17, \, a_n = 18 + (n-1) \cdot (-7)}
\]

#### (d) Sequence: \( 3, 3 \frac{1}{2}, 4, \ldots \) (16th term)

1. Identify \( a_1 \), \( d \), and \( n \):
- First term: \( a_1 = 3 \)
- Common difference: \( d = 3 \frac{1}{2} - 3 = \frac{7}{2} - 3 = \frac{7}{2} - \frac{6}{2} = \frac{1}{2} \)
- Position: \( n = 16 \)

2. Find the 16th term (\( a_{16} \)):
\[
a_n = a_1 + (n-1)d
\]
\[
a_{16} = 3 + (16-1) \cdot \frac{1}{2} = 3 + 15 \cdot \frac{1}{2} = 3 + \frac{15}{2} = 3 + 7.5 = 10.5
\]

3. Formula for the nth term:
\[
a_n = 3 + (n-1) \cdot \frac{1}{2}
\]

Answer for (d):
\[
\boxed{10.5, \, a_n = 3 + (n-1) \cdot \frac{1}{2}}
\]

---

Problem 3: Find the sum of the following arithmetic series and write in summation notation

#### (a) Series: \( 4, 11, \ldots \) to 16 terms

1. Identify \( a_1 \), \( d \), and \( n \):
- First term: \( a_1 = 4 \)
- Common difference: \( d = 11 - 4 = 7 \)
- Number of terms: \( n = 16 \)

2. Find the 16th term (\( a_{16} \)):
\[
a_n = a_1 + (n-1)d
\]
\[
a_{16} = 4 + (16-1) \cdot 7 = 4 + 15 \cdot 7 = 4 + 105 = 109
\]

3. Sum of the first 16 terms (\( S_{16} \)):
\[
S_n = \frac{n}{2} (a_1 + a_n)
\]
\[
S_{16} = \frac{16}{2} (4 + 109) = 8 \cdot 113 = 904
\]

4. Summation notation:
\[
S_{16} = \sum_{k=1}^{16} \left( 4 + (k-1) \cdot 7 \right)
\]

Answer for (a):
\[
\boxed{904, \, \sum_{k=1}^{16} \left( 4 + (k-1) \cdot 7 \right)}
\]

#### (b) Series: \( 19, 13, \ldots \) to 10 terms

1. Identify \( a_1 \), \( d \), and \( n \):
- First term: \( a_1 = 19 \)
- Common difference: \( d = 13 - 19 = -6 \)
- Number of terms: \( n = 10 \)

2. Find the 10th term (\( a_{10} \)):
\[
a_n = a_1 + (n-1)d
\]
\[
a_{10} = 19 + (10-1) \cdot (-6) = 19 + 9 \cdot (-6) = 19 - 54 = -35
\]

3. Sum of the first 10 terms (\( S_{10} \)):
\[
S_n = \frac{n}{2} (a_1 + a_n)
\]
\[
S_{10} = \frac{10}{2} (19 + (-35)) = 5 \cdot (-16) = -80
\]

4. Summation notation:
\[
S_{10} = \sum_{k=1}^{10} \left( 19 + (k-1) \cdot (-6) \right)
\]

Answer for (b):
\[
\boxed{-80, \, \sum_{k=1}^{10} \left( 19 + (k-1) \cdot (-6) \right)}
\]

#### (c) Series: \( 3, 8 \frac{1}{2}, \ldots \) to 20 terms

1. Identify \( a_1 \), \( d \), and \( n \):
- First term: \( a_1 = 3 \)
- Common difference: \( d = 8 \frac{1}{2} - 3 = \frac{17}{2} - 3 = \frac{17}{2} - \frac{6}{2} = \frac{11}{2} \)
- Number of terms: \( n = 20 \)

2. Find the 20th term (\( a_{20} \)):
\[
a_n = a_1 + (n-1)d
\]
\[
a_{20} = 3 + (20-1) \cdot \frac{11}{2} = 3 + 19 \cdot \frac{11}{2} = 3 + \frac{209}{2} = \frac{6}{2} + \frac{209}{2} = \frac{215}{2} = 107.5
\]

3. Sum of the first 20 terms (\( S_{20} \)):
\[
S_n = \frac{n}{2} (a_1 + a_n)
\]
\[
S_{20} = \frac{20}{2} \left( 3 + 107.5 \right) = 10 \cdot 110.5 = 1105
\]

4. Summation notation:
\[
S_{20} = \sum_{k=1}^{20} \left( 3 + (k-1) \cdot \frac{11}{2} \right)
\]

Answer for (c):
\[
\boxed{1105, \, \sum_{k=1}^{20} \left( 3 + (k-1) \cdot \frac{11}{2} \right)}
\]

#### (d) Series: \( -9, -1, \ldots \) to 8 terms

1. Identify \( a_1 \), \( d \), and \( n \):
- First term: \( a_1 = -9 \)
- Common difference: \( d = -1 - (-9) = 8 \)
- Number of terms: \( n = 8 \)

2. Find the 8th term (\( a_8 \)):
\[
a_n = a_1 + (n-1)d
\]
\[
a_8 = -9 + (8-1) \cdot 8 = -9 + 7 \cdot 8 = -9 + 56 = 47
\]

3. Sum of the first 8 terms (\( S_8 \)):
\[
S_n = \frac{n}{2} (a_1 + a_n)
\]
\[
S_8 = \frac{8}{2} (-9 + 47) = 4 \cdot 38 = 152
\]

4. Summation notation:
\[
S_8 = \sum_{k=1}^{8} \left( -9 + (k-1) \cdot 8 \right)
\]

Answer for (d):
\[
\boxed{152, \, \sum_{k=1}^{8} \left( -9 + (k-1) \cdot 8 \right)}
\]

#### (e) Series: \( 5, 9, 13, \ldots, 101 \)

1. Identify \( a_1 \), \( d \), and \( a_n \):
- First term: \( a_1 = 5 \)
- Common difference: \( d = 9 - 5 = 4 \)
- Last term: \( a_n = 101 \)

2. Find the number of terms (\( n \)):
\[
a_n = a_1 + (n-1)d
\]
\[
101 = 5 + (n-1) \cdot 4
\]
\[
101 = 5 + 4n - 4
\]
\[
101 = 1 + 4n
\]
\[
100 = 4n
\]
\[
n = 25
\]

3. Sum of the first 25 terms (\( S_{25} \)):
\[
S_n = \frac{n}{2} (a_1 + a_n)
\]
\[
S_{25} = \frac{25}{2} (5 + 101) = \frac{25}{2} \cdot 106 = 25 \cdot 53 = 1325
\]

4. Summation notation:
\[
S_{25} = \sum_{k=1}^{25} \left( 5 + (k-1) \cdot 4 \right)
\]

Answer for (e):
\[
\boxed{1325, \, \sum_{k=1}^{25} \left( 5 + (k-1) \cdot 4 \right)}
\]

#### (f) Series: \( 83, 80, 77, \ldots, 5 \)

1. Identify \( a_1 \), \( d \), and \( a_n \):
- First term: \( a_1 = 83 \)
- Common difference: \( d = 80 - 83 = -3 \)
- Last term: \( a_n = 5 \)

2. Find the number of terms (\( n \)):
\[
a_n = a_1 + (n-1)d
\]
\[
5 = 83 + (n-1) \cdot (-3)
\]
\[
5 = 83 - 3(n-1)
\]
\[
5 = 83 - 3n + 3
\]
\[
5 = 86 - 3n
\]
\[
3n = 86 - 5
\]
\[
3n = 81
\]
\[
n = 27
\]

3. Sum of the first 27 terms (\( S_{27} \)):
\[
S_n = \frac{n}{2} (a_1 + a_n)
\]
\[
S_{27} = \frac{27}{2} (83 + 5) = \frac{27}{2} \cdot 88 = 27 \cdot 44 = 1188
\]

4. Summation notation:
\[
S_{27} = \sum_{k=1}^{27} \left( 83 + (k-1) \cdot (-3) \right)
\]

Answer for (f):
\[
\boxed{1188, \, \sum_{k=1}^{27} \left( 83 + (k-1) \cdot (-3) \right)}
\]

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Final Answers:

1. (a) \(\boxed{31, \, a_n = 7 + (n-1) \cdot 4}\)
(b) \(\boxed{37, \, a_n = -7 + (n-1) \cdot 2}\)
(c) \(\boxed{-17, \, a_n = 18 + (n-1) \cdot (-7)}\)
(d) \(\boxed{10.5, \, a_n = 3 + (n-1) \cdot \frac{1}{2}}\)

3. (a) \(\boxed{904, \, \sum_{k=1}^{16} \left( 4 + (k-1) \cdot 7 \right)}\)
(b) \(\boxed{-80, \, \sum_{k=1}^{10} \left( 19 + (k-1) \cdot (-6) \right)}\)
(c) \(\boxed{1105, \, \sum_{k=1}^{20} \left( 3 + (k-1) \cdot \frac{11}{2} \right)}\)
(d) \(\boxed{152, \, \sum_{k=1}^{8} \left( -9 + (k-1) \cdot 8 \right)}\)
(e) \(\boxed{1325, \, \sum_{k=1}^{25} \left( 5 + (k-1) \cdot 4 \right)}\)
(f) \(\boxed{1188, \, \sum_{k=1}^{27} \left( 83 + (k-1) \cdot (-3) \right)}\)