Arithmetic Sequence — Sum of Even and Odd Terms — Dynamic Practice
Arithmetic Sequence — Sum of Even and Odd Terms — Dynamic Practice. Practice questions to deepen understanding of finding sums of terms at even and odd positions in an arithmetic sequence. Online math practice with full solutions and step-by-step explanations.
Dynamic practice in summing terms at even or odd positions of an arithmetic sequence — these form sub-sequences with a new common difference of 2d. New questions every attempt.
Question 1
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 15 terms, where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 15 terms, where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 256
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 256
Answer: 256
Question 2
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 17 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 17 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 8
Sum = 136
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 8
Sum = 136
Answer: 136
Question 3
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 3\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 3\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 7
Sum = 161
Terms at positions odd form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 7
Sum = 161
Answer: 161
Question 4
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 8
Sum = 128
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 8
Sum = 128
Answer: 128
Question 5
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 6
Sum = 60
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 6
Sum = 60
Answer: 60
Question 6
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 19 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 19 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 9
Sum = 144
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 9
Sum = 144
Answer: 144
Question 7
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 17 terms, where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 17 terms, where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 320
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 320
Answer: 320
Question 8
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 10 terms, where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 10 terms, where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 5
Sum = 120
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 5
Sum = 120
Answer: 120
Question 9
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 15 terms, where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 2\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 15 terms, where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 2\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 7
Sum = 154
Terms at positions even form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 7
Sum = 154
Answer: 154
Question 10
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 11 terms, where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 3\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 11 terms, where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 3\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 6
Sum = 138
Terms at positions odd form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 6
Sum = 138
Answer: 138
Question 11
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 6
Sum = 186
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 6
Sum = 186
Answer: 186
Question 12
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 19 terms, where:
• First term: \(a_1 = 6\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 19 terms, where:
• First term: \(a_1 = 6\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 10
Sum = 240
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 10
Sum = 240
Answer: 240
Question 13
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 18 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 18 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 9
Sum = 99
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 9
Sum = 99
Answer: 99
Question 14
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 7
Sum = 210
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 7
Sum = 210
Answer: 210
Question 15
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 2\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 2\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 6
Sum = 126
Terms at positions even form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 6
Sum = 126
Answer: 126
Question 16
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 6
Sum = 138
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 6
Sum = 138
Answer: 138
Question 17
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 6
Sum = 150
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 6
Sum = 150
Answer: 150
Question 18
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 6
Sum = 132
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 6
Sum = 132
Answer: 132
Question 19
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 10 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 10 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 5
Sum = 65
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 5
Sum = 65
Answer: 65
Question 20
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 6
Sum = 114
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 6
Sum = 114
Answer: 114
Question 21
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 11 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 11 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 5
Sum = 120
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 5
Sum = 120
Answer: 120
Question 22
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 10 terms, where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 10 terms, where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 5
Sum = 100
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 5
Sum = 100
Answer: 100
Question 23
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 7
Sum = 84
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 7
Sum = 84
Answer: 84
Question 24
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 7
Sum = 105
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 7
Sum = 105
Answer: 105
Question 25
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 3\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 3\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 7
Sum = 63
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 7
Sum = 63
Answer: 63
Question 26
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 17 terms, where:
• First term: \(a_1 = 3\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 17 terms, where:
• First term: \(a_1 = 3\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 280
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 280
Answer: 280
Question 27
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 7
Sum = 147
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 7
Sum = 147
Answer: 147
Question 28
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 1\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 1\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 264
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 264
Answer: 264
Question 29
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 10 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 10 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 5
Sum = 35
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 5
Sum = 35
Answer: 35
Question 30
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 17 terms, where:
• First term: \(a_1 = 1\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 17 terms, where:
• First term: \(a_1 = 1\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 264
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 264
Answer: 264
Question 31
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 19 terms, where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 19 terms, where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 10
Sum = 440
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 10
Sum = 440
Answer: 440
Question 32
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 3\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 3\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 7
Sum = 105
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 7
Sum = 105
Answer: 105
Question 33
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 7
Sum = 77
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 7
Sum = 77
Answer: 77
Question 34
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 10 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 10 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 5
Sum = 65
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 5
Sum = 65
Answer: 65
Question 35
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 15 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 15 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 8
Sum = 168
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 8
Sum = 168
Answer: 168
Question 36
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 11 terms, where:
• First term: \(a_1 = 1\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 11 terms, where:
• First term: \(a_1 = 1\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 5
Sum = 80
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 5
Sum = 80
Answer: 80
Question 37
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 6
Sum = 150
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 6
Sum = 150
Answer: 150
Question 38
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 18 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 18 terms, where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 9
Sum = 333
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 9
Sum = 333
Answer: 333
Question 39
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 11 terms, where:
• First term: \(a_1 = 6\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 11 terms, where:
• First term: \(a_1 = 6\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 5
Sum = 105
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 5
Sum = 105
Answer: 105
Question 40
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 3\)
Find the sum of terms at even positions.
Explanation:
Solution – Arithmetic Sequence:
📝 Important formulas:
\(a_n = a_1 + (n-1) \cdot d\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
\(S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(2a_1 + (n-1)d)}{2}\)
🔢 Solution:
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 8
Sum = 248
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 8
Sum = 248
Answer: 248