Arithmetic Sequence — Sum at Even or Odd Positions — Dynamic Practice
Arithmetic Sequence — Sum at Even or Odd Positions — Dynamic Practice. Practice questions to deepen understanding of summing terms at even or odd positions in an arithmetic sequence. Online math practice with full solutions and detailed explanations.
Dynamic practice in summing terms at even or odd positions — they form sub-sequences with new common difference 2d. New questions every attempt.
Question 1
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 2\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 2\)
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 = 114
Terms at positions even form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 6
Sum = 114
Answer: 114
Question 2
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 15 terms, where:
• First term: \(a_1 = 6\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 15 terms, where:
• First term: \(a_1 = 6\)
• 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 = 272
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 272
Answer: 272
Question 3
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 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 13 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: 7
Sum = 224
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 7
Sum = 224
Answer: 224
Question 4
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 17 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 3\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 17 terms, where:
• First term: \(a_1 = 2\)
• 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: 9
Sum = 234
Terms at positions odd form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 9
Sum = 234
Answer: 234
Question 5
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 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 13 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: 6
Sum = 150
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 6
Sum = 150
Answer: 150
Question 6
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 17 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 17 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: 8
Sum = 224
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 8
Sum = 224
Answer: 224
Question 7
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 12 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 12 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: 6
Sum = 156
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 6
Sum = 156
Answer: 156
Question 8
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 2\)
• 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 = 128
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 8
Sum = 128
Answer: 128
Question 9
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 1\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 1\)
• 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 = 120
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 8
Sum = 120
Answer: 120
Question 10
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 10 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 10 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: 5
Sum = 105
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 5
Sum = 105
Answer: 105
Question 11
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 6\)
• The common difference: \(d = 1\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 6\)
• 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: 6
Sum = 72
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 6
Sum = 72
Answer: 72
Question 12
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 3\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 7\)
• 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 = 175
Terms at positions odd form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 7
Sum = 175
Answer: 175
Question 13
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 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 14 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 14
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 7\)
• 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 = 91
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 7
Sum = 91
Answer: 91
Question 15
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 12 terms, where:
• First term: \(a_1 = 2\)
• 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 = 72
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 6
Sum = 72
Answer: 72
Question 16
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 even 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 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: 6
Sum = 90
Terms at positions even form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 6
Sum = 90
Answer: 90
Question 17
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 3\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 3\)
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 = 189
Terms at positions odd form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 7
Sum = 189
Answer: 189
Question 18
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 even 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 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 = 115
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 5
Sum = 115
Answer: 115
Question 19
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 16 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 16 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: 8
Sum = 152
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 8
Sum = 152
Answer: 152
Question 20
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 18 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 18 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: 9
Sum = 99
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 9
Sum = 99
Answer: 99
Question 21
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 22
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 18 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 18 terms, where:
• First term: \(a_1 = 9\)
• 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 = 369
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 9
Sum = 369
Answer: 369
Question 23
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 19 terms, where:
• First term: \(a_1 = 6\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 19 terms, where:
• First term: \(a_1 = 6\)
• 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: 9
Sum = 378
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 9
Sum = 378
Answer: 378
Question 24
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 12 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 12 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: 6
Sum = 114
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 6
Sum = 114
Answer: 114
Question 25
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 odd 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 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 = 315
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 9
Sum = 315
Answer: 315
Question 26
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 2\)
• 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 = 98
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 7
Sum = 98
Answer: 98
Question 27
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 19 terms, where:
• First term: \(a_1 = 9\)
• The common difference: \(d = 4\)
Find the sum of terms at even positions.
Given an arithmetic sequence with 19 terms, where:
• First term: \(a_1 = 9\)
• 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: 9
Sum = 405
Terms at positions even form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 9
Sum = 405
Answer: 405
Question 28
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 11 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 11 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: 5
Sum = 95
Terms at positions even form a new arithmetic sequence with difference \(2d = 6\)
Number of terms: 5
Sum = 95
Answer: 95
Question 29
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 17 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 17 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: 9
Sum = 99
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 9
Sum = 99
Answer: 99
Question 30
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 10 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 10 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 31
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 17 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 17 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: 9
Sum = 207
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 9
Sum = 207
Answer: 207
Question 32
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 15 terms, where:
• First term: \(a_1 = 2\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 15 terms, where:
• First term: \(a_1 = 2\)
• 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 = 240
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 8
Sum = 240
Answer: 240
Question 33
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 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 14 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: 7
Sum = 196
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 7
Sum = 196
Answer: 196
Question 34
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 1\)
• The common difference: \(d = 4\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 13 terms, where:
• First term: \(a_1 = 1\)
• 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: 7
Sum = 175
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 7
Sum = 175
Answer: 175
Question 35
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 odd 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 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 36
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 37
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 17 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 17 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: 9
Sum = 324
Terms at positions odd form a new arithmetic sequence with difference \(2d = 8\)
Number of terms: 9
Sum = 324
Answer: 324
Question 38
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 1\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 14 terms, where:
• First term: \(a_1 = 8\)
• 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 = 98
Terms at positions odd form a new arithmetic sequence with difference \(2d = 2\)
Number of terms: 7
Sum = 98
Answer: 98
Question 39
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 2\)
Find the sum of terms at odd positions.
Given an arithmetic sequence with 16 terms, where:
• First term: \(a_1 = 8\)
• 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 = 176
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 8
Sum = 176
Answer: 176
Question 40
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence with 12 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 12 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: 6
Sum = 96
Terms at positions odd form a new arithmetic sequence with difference \(2d = 4\)
Number of terms: 6
Sum = 96
Answer: 96