Arithmetic Sequence — Finding the Difference d — Part 2 — Dynamic Practice
Arithmetic Sequence — Finding the Difference d — Part 2 — Dynamic Practice. Practice questions to deepen understanding of finding the common difference d in an arithmetic sequence — advanced variations. Online math practice with full solutions and step-by-step explanations.
Dynamic practice in finding d from given data using the formula aₙ = a₁ + (n−1)d. New questions every attempt.
Question 1
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The 11-th term: \(a_{11} = 16\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The 11-th term: \(a_{11} = 16\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{16 - -4}{11 - 1} = \frac{20}{10} = 2\)
Answer: 2
Question 2
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 1\)
• The 5-th term: \(a_{5} = -3\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 1\)
• The 5-th term: \(a_{5} = -3\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-3 - 1}{5 - 1} = \frac{-4}{4} = -1\)
Answer: -1
Question 3
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 13\)
• The 9-th term: \(a_{9} = 69\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 13\)
• The 9-th term: \(a_{9} = 69\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{69 - 13}{9 - 1} = \frac{56}{8} = 7\)
Answer: 7
Question 4
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 14\)
• The 9-th term: \(a_{9} = -26\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 14\)
• The 9-th term: \(a_{9} = -26\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-26 - 14}{9 - 1} = \frac{-40}{8} = -5\)
Answer: -5
Question 5
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The 6-th term: \(a_{6} = 18\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The 6-th term: \(a_{6} = 18\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{18 - 3}{6 - 1} = \frac{15}{5} = 3\)
Answer: 3
Question 6
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 12\)
• The 8-th term: \(a_{8} = 68\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 12\)
• The 8-th term: \(a_{8} = 68\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{68 - 12}{8 - 1} = \frac{56}{7} = 8\)
Answer: 8
Question 7
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The 10-th term: \(a_{10} = 66\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The 10-th term: \(a_{10} = 66\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{66 - 3}{10 - 1} = \frac{63}{9} = 7\)
Answer: 7
Question 8
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -2\)
• The 6-th term: \(a_{6} = -27\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = -2\)
• The 6-th term: \(a_{6} = -27\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-27 - -2}{6 - 1} = \frac{-25}{5} = -5\)
Answer: -5
Question 9
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 8-th term: \(a_{8} = 55\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 8-th term: \(a_{8} = 55\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{55 - 6}{8 - 1} = \frac{49}{7} = 7\)
Answer: 7
Question 10
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 5-th term: \(a_{5} = 10\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 5-th term: \(a_{5} = 10\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{10 - 6}{5 - 1} = \frac{4}{4} = 1\)
Answer: 1
Question 11
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The 8-th term: \(a_{8} = -24\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The 8-th term: \(a_{8} = -24\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-24 - 4}{8 - 1} = \frac{-28}{7} = -4\)
Answer: -4
Question 12
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 2\)
• The 9-th term: \(a_{9} = -30\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 2\)
• The 9-th term: \(a_{9} = -30\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-30 - 2}{9 - 1} = \frac{-32}{8} = -4\)
Answer: -4
Question 13
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -1\)
• The 11-th term: \(a_{11} = 79\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = -1\)
• The 11-th term: \(a_{11} = 79\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{79 - -1}{11 - 1} = \frac{80}{10} = 8\)
Answer: 8
Question 14
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 10-th term: \(a_{10} = -3\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 10-th term: \(a_{10} = -3\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-3 - 6}{10 - 1} = \frac{-9}{9} = -1\)
Answer: -1
Question 15
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 7\)
• The 6-th term: \(a_{6} = -13\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 7\)
• The 6-th term: \(a_{6} = -13\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-13 - 7}{6 - 1} = \frac{-20}{5} = -4\)
Answer: -4
Question 16
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 10-th term: \(a_{10} = 60\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 10-th term: \(a_{10} = 60\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{60 - 6}{10 - 1} = \frac{54}{9} = 6\)
Answer: 6
Question 17
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 2\)
• The 9-th term: \(a_{9} = 66\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 2\)
• The 9-th term: \(a_{9} = 66\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{66 - 2}{9 - 1} = \frac{64}{8} = 8\)
Answer: 8
Question 18
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 14\)
• The 6-th term: \(a_{6} = 39\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 14\)
• The 6-th term: \(a_{6} = 39\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{39 - 14}{6 - 1} = \frac{25}{5} = 5\)
Answer: 5
Question 19
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 13\)
• The 5-th term: \(a_{5} = 45\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 13\)
• The 5-th term: \(a_{5} = 45\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{45 - 13}{5 - 1} = \frac{32}{4} = 8\)
Answer: 8
Question 20
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 5\)
• The 9-th term: \(a_{9} = 21\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 5\)
• The 9-th term: \(a_{9} = 21\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{21 - 5}{9 - 1} = \frac{16}{8} = 2\)
Answer: 2
Question 21
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 8\)
• The 8-th term: \(a_{8} = 50\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 8\)
• The 8-th term: \(a_{8} = 50\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{50 - 8}{8 - 1} = \frac{42}{7} = 6\)
Answer: 6
Question 22
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The 8-th term: \(a_{8} = -39\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The 8-th term: \(a_{8} = -39\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-39 - -4}{8 - 1} = \frac{-35}{7} = -5\)
Answer: -5
Question 23
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 8-th term: \(a_{8} = -22\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 8-th term: \(a_{8} = -22\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-22 - 6}{8 - 1} = \frac{-28}{7} = -4\)
Answer: -4
Question 24
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 10\)
• The 7-th term: \(a_{7} = 4\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 10\)
• The 7-th term: \(a_{7} = 4\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{4 - 10}{7 - 1} = \frac{-6}{6} = -1\)
Answer: -1
Question 25
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 7\)
• The 5-th term: \(a_{5} = 27\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 7\)
• The 5-th term: \(a_{5} = 27\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{27 - 7}{5 - 1} = \frac{20}{4} = 5\)
Answer: 5
Question 26
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 10-th term: \(a_{10} = -39\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The 10-th term: \(a_{10} = -39\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-39 - 6}{10 - 1} = \frac{-45}{9} = -5\)
Answer: -5
Question 27
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 7\)
• The 8-th term: \(a_{8} = -21\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 7\)
• The 8-th term: \(a_{8} = -21\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-21 - 7}{8 - 1} = \frac{-28}{7} = -4\)
Answer: -4
Question 28
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -1\)
• The 10-th term: \(a_{10} = 62\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = -1\)
• The 10-th term: \(a_{10} = 62\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{62 - -1}{10 - 1} = \frac{63}{9} = 7\)
Answer: 7
Question 29
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 0\)
• The 10-th term: \(a_{10} = 63\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 0\)
• The 10-th term: \(a_{10} = 63\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{63 - 0}{10 - 1} = \frac{63}{9} = 7\)
Answer: 7
Question 30
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 13\)
• The 10-th term: \(a_{10} = 31\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 13\)
• The 10-th term: \(a_{10} = 31\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{31 - 13}{10 - 1} = \frac{18}{9} = 2\)
Answer: 2
Question 31
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 10\)
• The 8-th term: \(a_{8} = 66\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 10\)
• The 8-th term: \(a_{8} = 66\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{66 - 10}{8 - 1} = \frac{56}{7} = 8\)
Answer: 8
Question 32
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -3\)
• The 5-th term: \(a_{5} = 13\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = -3\)
• The 5-th term: \(a_{5} = 13\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{13 - -3}{5 - 1} = \frac{16}{4} = 4\)
Answer: 4
Question 33
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The 8-th term: \(a_{8} = 52\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The 8-th term: \(a_{8} = 52\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{52 - 3}{8 - 1} = \frac{49}{7} = 7\)
Answer: 7
Question 34
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -5\)
• The 5-th term: \(a_{5} = -13\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = -5\)
• The 5-th term: \(a_{5} = -13\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-13 - -5}{5 - 1} = \frac{-8}{4} = -2\)
Answer: -2
Question 35
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The 8-th term: \(a_{8} = -18\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The 8-th term: \(a_{8} = -18\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{-18 - 3}{8 - 1} = \frac{-21}{7} = -3\)
Answer: -3
Question 36
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The 7-th term: \(a_{7} = 9\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The 7-th term: \(a_{7} = 9\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{9 - 3}{7 - 1} = \frac{6}{6} = 1\)
Answer: 1
Question 37
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 0\)
• The 8-th term: \(a_{8} = 7\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 0\)
• The 8-th term: \(a_{8} = 7\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{7 - 0}{8 - 1} = \frac{7}{7} = 1\)
Answer: 1
Question 38
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 0\)
• The 11-th term: \(a_{11} = 20\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 0\)
• The 11-th term: \(a_{11} = 20\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{20 - 0}{11 - 1} = \frac{20}{10} = 2\)
Answer: 2
Question 39
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The 6-th term: \(a_{6} = 39\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The 6-th term: \(a_{6} = 39\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{39 - 4}{6 - 1} = \frac{35}{5} = 7\)
Answer: 7
Question 40
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 0\)
• The 9-th term: \(a_{9} = 48\)
Find the common difference \(d\).
Given an arithmetic sequence where:
• First term: \(a_1 = 0\)
• The 9-th term: \(a_{9} = 48\)
Find the common difference \(d\).
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:
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
We will use the formula: \(d = \frac{a_n - a_1}{n - 1}\)
\(d = \frac{48 - 0}{9 - 1} = \frac{48}{8} = 6\)
Answer: 6