Arithmetic Sequence — Finding the General Term aₙ — Dynamic Practice
Arithmetic Sequence — Finding the General Term aₙ — Dynamic Practice. Practice questions to deepen understanding of finding the value of the general term aₙ in an arithmetic sequence. Online math practice with full solutions and detailed explanations.
Dynamic practice in calculating aₙ — given a₁ and d, compute the value of a specific term. New questions every attempt.
Question 1
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 11\)
• The common difference: \(d = -3\)
Find the 18-th term (i.e. \(a_{18}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 11\)
• The common difference: \(d = -3\)
Find the 18-th term (i.e. \(a_{18}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{18} = 11 + (18-1) \cdot -3\)
\(a_{18} = 11 + 17 \cdot -3\)
\(a_{18} = 11 + -51 = -40\)
\(a_{18} = 11 + 17 \cdot -3\)
\(a_{18} = 11 + -51 = -40\)
Answer: -40
Question 2
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 9\)
• The common difference: \(d = -3\)
Find the 10-th term (i.e. \(a_{10}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 9\)
• The common difference: \(d = -3\)
Find the 10-th term (i.e. \(a_{10}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{10} = 9 + (10-1) \cdot -3\)
\(a_{10} = 9 + 9 \cdot -3\)
\(a_{10} = 9 + -27 = -18\)
\(a_{10} = 9 + 9 \cdot -3\)
\(a_{10} = 9 + -27 = -18\)
Answer: -18
Question 3
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 7\)
• The common difference: \(d = -4\)
Find the 19-th term (i.e. \(a_{19}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 7\)
• The common difference: \(d = -4\)
Find the 19-th term (i.e. \(a_{19}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{19} = 7 + (19-1) \cdot -4\)
\(a_{19} = 7 + 18 \cdot -4\)
\(a_{19} = 7 + -72 = -65\)
\(a_{19} = 7 + 18 \cdot -4\)
\(a_{19} = 7 + -72 = -65\)
Answer: -65
Question 4
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 1\)
• The common difference: \(d = 5\)
Find the 11-th term (i.e. \(a_{11}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 1\)
• The common difference: \(d = 5\)
Find the 11-th term (i.e. \(a_{11}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{11} = 1 + (11-1) \cdot 5\)
\(a_{11} = 1 + 10 \cdot 5\)
\(a_{11} = 1 + 50 = 51\)
\(a_{11} = 1 + 10 \cdot 5\)
\(a_{11} = 1 + 50 = 51\)
Answer: 51
Question 5
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 11\)
• The common difference: \(d = -3\)
Find the 19-th term (i.e. \(a_{19}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 11\)
• The common difference: \(d = -3\)
Find the 19-th term (i.e. \(a_{19}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{19} = 11 + (19-1) \cdot -3\)
\(a_{19} = 11 + 18 \cdot -3\)
\(a_{19} = 11 + -54 = -43\)
\(a_{19} = 11 + 18 \cdot -3\)
\(a_{19} = 11 + -54 = -43\)
Answer: -43
Question 6
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 14\)
• The common difference: \(d = 2\)
Find the 8-th term (i.e. \(a_{8}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 14\)
• The common difference: \(d = 2\)
Find the 8-th term (i.e. \(a_{8}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{8} = 14 + (8-1) \cdot 2\)
\(a_{8} = 14 + 7 \cdot 2\)
\(a_{8} = 14 + 14 = 28\)
\(a_{8} = 14 + 7 \cdot 2\)
\(a_{8} = 14 + 14 = 28\)
Answer: 28
Question 7
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 0\)
• The common difference: \(d = 3\)
Find the 8-th term (i.e. \(a_{8}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 0\)
• The common difference: \(d = 3\)
Find the 8-th term (i.e. \(a_{8}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{8} = 0 + (8-1) \cdot 3\)
\(a_{8} = 0 + 7 \cdot 3\)
\(a_{8} = 0 + 21 = 21\)
\(a_{8} = 0 + 7 \cdot 3\)
\(a_{8} = 0 + 21 = 21\)
Answer: 21
Question 8
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 14\)
• The common difference: \(d = -2\)
Find the 15-th term (i.e. \(a_{15}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 14\)
• The common difference: \(d = -2\)
Find the 15-th term (i.e. \(a_{15}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{15} = 14 + (15-1) \cdot -2\)
\(a_{15} = 14 + 14 \cdot -2\)
\(a_{15} = 14 + -28 = -14\)
\(a_{15} = 14 + 14 \cdot -2\)
\(a_{15} = 14 + -28 = -14\)
Answer: -14
Question 9
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 1\)
• The common difference: \(d = -4\)
Find the 14-th term (i.e. \(a_{14}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 1\)
• The common difference: \(d = -4\)
Find the 14-th term (i.e. \(a_{14}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{14} = 1 + (14-1) \cdot -4\)
\(a_{14} = 1 + 13 \cdot -4\)
\(a_{14} = 1 + -52 = -51\)
\(a_{14} = 1 + 13 \cdot -4\)
\(a_{14} = 1 + -52 = -51\)
Answer: -51
Question 10
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 13\)
• The common difference: \(d = 4\)
Find the 19-th term (i.e. \(a_{19}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 13\)
• The common difference: \(d = 4\)
Find the 19-th term (i.e. \(a_{19}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{19} = 13 + (19-1) \cdot 4\)
\(a_{19} = 13 + 18 \cdot 4\)
\(a_{19} = 13 + 72 = 85\)
\(a_{19} = 13 + 18 \cdot 4\)
\(a_{19} = 13 + 72 = 85\)
Answer: 85
Question 11
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 14\)
• The common difference: \(d = -4\)
Find the 17-th term (i.e. \(a_{17}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 14\)
• The common difference: \(d = -4\)
Find the 17-th term (i.e. \(a_{17}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{17} = 14 + (17-1) \cdot -4\)
\(a_{17} = 14 + 16 \cdot -4\)
\(a_{17} = 14 + -64 = -50\)
\(a_{17} = 14 + 16 \cdot -4\)
\(a_{17} = 14 + -64 = -50\)
Answer: -50
Question 12
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 13\)
• The common difference: \(d = -1\)
Find the 12-th term (i.e. \(a_{12}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 13\)
• The common difference: \(d = -1\)
Find the 12-th term (i.e. \(a_{12}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{12} = 13 + (12-1) \cdot -1\)
\(a_{12} = 13 + 11 \cdot -1\)
\(a_{12} = 13 + -11 = 2\)
\(a_{12} = 13 + 11 \cdot -1\)
\(a_{12} = 13 + -11 = 2\)
Answer: 2
Question 13
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -1\)
• The common difference: \(d = -5\)
Find the 11-th term (i.e. \(a_{11}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -1\)
• The common difference: \(d = -5\)
Find the 11-th term (i.e. \(a_{11}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{11} = -1 + (11-1) \cdot -5\)
\(a_{11} = -1 + 10 \cdot -5\)
\(a_{11} = -1 + -50 = -51\)
\(a_{11} = -1 + 10 \cdot -5\)
\(a_{11} = -1 + -50 = -51\)
Answer: -51
Question 14
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The common difference: \(d = 7\)
Find the 14-th term (i.e. \(a_{14}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The common difference: \(d = 7\)
Find the 14-th term (i.e. \(a_{14}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{14} = 3 + (14-1) \cdot 7\)
\(a_{14} = 3 + 13 \cdot 7\)
\(a_{14} = 3 + 91 = 94\)
\(a_{14} = 3 + 13 \cdot 7\)
\(a_{14} = 3 + 91 = 94\)
Answer: 94
Question 15
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -1\)
• The common difference: \(d = 4\)
Find the 15-th term (i.e. \(a_{15}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -1\)
• The common difference: \(d = 4\)
Find the 15-th term (i.e. \(a_{15}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{15} = -1 + (15-1) \cdot 4\)
\(a_{15} = -1 + 14 \cdot 4\)
\(a_{15} = -1 + 56 = 55\)
\(a_{15} = -1 + 14 \cdot 4\)
\(a_{15} = -1 + 56 = 55\)
Answer: 55
Question 16
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 11\)
• The common difference: \(d = -2\)
Find the 9-th term (i.e. \(a_{9}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 11\)
• The common difference: \(d = -2\)
Find the 9-th term (i.e. \(a_{9}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{9} = 11 + (9-1) \cdot -2\)
\(a_{9} = 11 + 8 \cdot -2\)
\(a_{9} = 11 + -16 = -5\)
\(a_{9} = 11 + 8 \cdot -2\)
\(a_{9} = 11 + -16 = -5\)
Answer: -5
Question 17
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The common difference: \(d = 4\)
Find the 16-th term (i.e. \(a_{16}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The common difference: \(d = 4\)
Find the 16-th term (i.e. \(a_{16}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{16} = 3 + (16-1) \cdot 4\)
\(a_{16} = 3 + 15 \cdot 4\)
\(a_{16} = 3 + 60 = 63\)
\(a_{16} = 3 + 15 \cdot 4\)
\(a_{16} = 3 + 60 = 63\)
Answer: 63
Question 18
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -1\)
• The common difference: \(d = 1\)
Find the 16-th term (i.e. \(a_{16}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -1\)
• The common difference: \(d = 1\)
Find the 16-th term (i.e. \(a_{16}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{16} = -1 + (16-1) \cdot 1\)
\(a_{16} = -1 + 15 \cdot 1\)
\(a_{16} = -1 + 15 = 14\)
\(a_{16} = -1 + 15 \cdot 1\)
\(a_{16} = -1 + 15 = 14\)
Answer: 14
Question 19
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = -2\)
Find the 15-th term (i.e. \(a_{15}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = -2\)
Find the 15-th term (i.e. \(a_{15}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{15} = -4 + (15-1) \cdot -2\)
\(a_{15} = -4 + 14 \cdot -2\)
\(a_{15} = -4 + -28 = -32\)
\(a_{15} = -4 + 14 \cdot -2\)
\(a_{15} = -4 + -28 = -32\)
Answer: -32
Question 20
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 4\)
Find the 19-th term (i.e. \(a_{19}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 4\)
Find the 19-th term (i.e. \(a_{19}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{19} = 4 + (19-1) \cdot 4\)
\(a_{19} = 4 + 18 \cdot 4\)
\(a_{19} = 4 + 72 = 76\)
\(a_{19} = 4 + 18 \cdot 4\)
\(a_{19} = 4 + 72 = 76\)
Answer: 76
Question 21
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = 7\)
Find the 17-th term (i.e. \(a_{17}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = 7\)
Find the 17-th term (i.e. \(a_{17}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{17} = -4 + (17-1) \cdot 7\)
\(a_{17} = -4 + 16 \cdot 7\)
\(a_{17} = -4 + 112 = 108\)
\(a_{17} = -4 + 16 \cdot 7\)
\(a_{17} = -4 + 112 = 108\)
Answer: 108
Question 22
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The common difference: \(d = -1\)
Find the 9-th term (i.e. \(a_{9}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The common difference: \(d = -1\)
Find the 9-th term (i.e. \(a_{9}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{9} = 4 + (9-1) \cdot -1\)
\(a_{9} = 4 + 8 \cdot -1\)
\(a_{9} = 4 + -8 = -4\)
\(a_{9} = 4 + 8 \cdot -1\)
\(a_{9} = 4 + -8 = -4\)
Answer: -4
Question 23
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The common difference: \(d = 3\)
Find the 16-th term (i.e. \(a_{16}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 6\)
• The common difference: \(d = 3\)
Find the 16-th term (i.e. \(a_{16}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{16} = 6 + (16-1) \cdot 3\)
\(a_{16} = 6 + 15 \cdot 3\)
\(a_{16} = 6 + 45 = 51\)
\(a_{16} = 6 + 15 \cdot 3\)
\(a_{16} = 6 + 45 = 51\)
Answer: 51
Question 24
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 7\)
Find the 9-th term (i.e. \(a_{9}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 7\)
Find the 9-th term (i.e. \(a_{9}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{9} = 4 + (9-1) \cdot 7\)
\(a_{9} = 4 + 8 \cdot 7\)
\(a_{9} = 4 + 56 = 60\)
\(a_{9} = 4 + 8 \cdot 7\)
\(a_{9} = 4 + 56 = 60\)
Answer: 60
Question 25
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 11\)
• The common difference: \(d = -1\)
Find the 15-th term (i.e. \(a_{15}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 11\)
• The common difference: \(d = -1\)
Find the 15-th term (i.e. \(a_{15}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{15} = 11 + (15-1) \cdot -1\)
\(a_{15} = 11 + 14 \cdot -1\)
\(a_{15} = 11 + -14 = -3\)
\(a_{15} = 11 + 14 \cdot -1\)
\(a_{15} = 11 + -14 = -3\)
Answer: -3
Question 26
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 11\)
• The common difference: \(d = -3\)
Find the 14-th term (i.e. \(a_{14}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 11\)
• The common difference: \(d = -3\)
Find the 14-th term (i.e. \(a_{14}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{14} = 11 + (14-1) \cdot -3\)
\(a_{14} = 11 + 13 \cdot -3\)
\(a_{14} = 11 + -39 = -28\)
\(a_{14} = 11 + 13 \cdot -3\)
\(a_{14} = 11 + -39 = -28\)
Answer: -28
Question 27
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = 3\)
Find the 9-th term (i.e. \(a_{9}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = 3\)
Find the 9-th term (i.e. \(a_{9}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{9} = -4 + (9-1) \cdot 3\)
\(a_{9} = -4 + 8 \cdot 3\)
\(a_{9} = -4 + 24 = 20\)
\(a_{9} = -4 + 8 \cdot 3\)
\(a_{9} = -4 + 24 = 20\)
Answer: 20
Question 28
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = -5\)
Find the 11-th term (i.e. \(a_{11}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = -5\)
Find the 11-th term (i.e. \(a_{11}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{11} = -4 + (11-1) \cdot -5\)
\(a_{11} = -4 + 10 \cdot -5\)
\(a_{11} = -4 + -50 = -54\)
\(a_{11} = -4 + 10 \cdot -5\)
\(a_{11} = -4 + -50 = -54\)
Answer: -54
Question 29
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = 5\)
Find the 11-th term (i.e. \(a_{11}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = 5\)
Find the 11-th term (i.e. \(a_{11}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{11} = -4 + (11-1) \cdot 5\)
\(a_{11} = -4 + 10 \cdot 5\)
\(a_{11} = -4 + 50 = 46\)
\(a_{11} = -4 + 10 \cdot 5\)
\(a_{11} = -4 + 50 = 46\)
Answer: 46
Question 30
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 6\)
Find the 18-th term (i.e. \(a_{18}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 6\)
Find the 18-th term (i.e. \(a_{18}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{18} = 8 + (18-1) \cdot 6\)
\(a_{18} = 8 + 17 \cdot 6\)
\(a_{18} = 8 + 102 = 110\)
\(a_{18} = 8 + 17 \cdot 6\)
\(a_{18} = 8 + 102 = 110\)
Answer: 110
Question 31
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The common difference: \(d = -5\)
Find the 16-th term (i.e. \(a_{16}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 3\)
• The common difference: \(d = -5\)
Find the 16-th term (i.e. \(a_{16}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{16} = 3 + (16-1) \cdot -5\)
\(a_{16} = 3 + 15 \cdot -5\)
\(a_{16} = 3 + -75 = -72\)
\(a_{16} = 3 + 15 \cdot -5\)
\(a_{16} = 3 + -75 = -72\)
Answer: -72
Question 32
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = 3\)
Find the 10-th term (i.e. \(a_{10}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -4\)
• The common difference: \(d = 3\)
Find the 10-th term (i.e. \(a_{10}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{10} = -4 + (10-1) \cdot 3\)
\(a_{10} = -4 + 9 \cdot 3\)
\(a_{10} = -4 + 27 = 23\)
\(a_{10} = -4 + 9 \cdot 3\)
\(a_{10} = -4 + 27 = 23\)
Answer: 23
Question 33
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -3\)
• The common difference: \(d = 7\)
Find the 10-th term (i.e. \(a_{10}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -3\)
• The common difference: \(d = 7\)
Find the 10-th term (i.e. \(a_{10}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{10} = -3 + (10-1) \cdot 7\)
\(a_{10} = -3 + 9 \cdot 7\)
\(a_{10} = -3 + 63 = 60\)
\(a_{10} = -3 + 9 \cdot 7\)
\(a_{10} = -3 + 63 = 60\)
Answer: 60
Question 34
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 8\)
Find the 19-th term (i.e. \(a_{19}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 5\)
• The common difference: \(d = 8\)
Find the 19-th term (i.e. \(a_{19}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{19} = 5 + (19-1) \cdot 8\)
\(a_{19} = 5 + 18 \cdot 8\)
\(a_{19} = 5 + 144 = 149\)
\(a_{19} = 5 + 18 \cdot 8\)
\(a_{19} = 5 + 144 = 149\)
Answer: 149
Question 35
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 5\)
• The common difference: \(d = -2\)
Find the 19-th term (i.e. \(a_{19}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 5\)
• The common difference: \(d = -2\)
Find the 19-th term (i.e. \(a_{19}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{19} = 5 + (19-1) \cdot -2\)
\(a_{19} = 5 + 18 \cdot -2\)
\(a_{19} = 5 + -36 = -31\)
\(a_{19} = 5 + 18 \cdot -2\)
\(a_{19} = 5 + -36 = -31\)
Answer: -31
Question 36
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 1\)
Find the 15-th term (i.e. \(a_{15}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 4\)
• The common difference: \(d = 1\)
Find the 15-th term (i.e. \(a_{15}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{15} = 4 + (15-1) \cdot 1\)
\(a_{15} = 4 + 14 \cdot 1\)
\(a_{15} = 4 + 14 = 18\)
\(a_{15} = 4 + 14 \cdot 1\)
\(a_{15} = 4 + 14 = 18\)
Answer: 18
Question 37
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = -2\)
• The common difference: \(d = -4\)
Find the 19-th term (i.e. \(a_{19}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = -2\)
• The common difference: \(d = -4\)
Find the 19-th term (i.e. \(a_{19}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{19} = -2 + (19-1) \cdot -4\)
\(a_{19} = -2 + 18 \cdot -4\)
\(a_{19} = -2 + -72 = -74\)
\(a_{19} = -2 + 18 \cdot -4\)
\(a_{19} = -2 + -72 = -74\)
Answer: -74
Question 38
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 7\)
Find the 11-th term (i.e. \(a_{11}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 7\)
• The common difference: \(d = 7\)
Find the 11-th term (i.e. \(a_{11}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{11} = 7 + (11-1) \cdot 7\)
\(a_{11} = 7 + 10 \cdot 7\)
\(a_{11} = 7 + 70 = 77\)
\(a_{11} = 7 + 10 \cdot 7\)
\(a_{11} = 7 + 70 = 77\)
Answer: 77
Question 39
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 3\)
Find the 14-th term (i.e. \(a_{14}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 8\)
• The common difference: \(d = 3\)
Find the 14-th term (i.e. \(a_{14}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{14} = 8 + (14-1) \cdot 3\)
\(a_{14} = 8 + 13 \cdot 3\)
\(a_{14} = 8 + 39 = 47\)
\(a_{14} = 8 + 13 \cdot 3\)
\(a_{14} = 8 + 39 = 47\)
Answer: 47
Question 40
2.50 pts
📊 Arithmetic Sequence:
Given an arithmetic sequence where:
• First term: \(a_1 = 10\)
• The common difference: \(d = 7\)
Find the 9-th term (i.e. \(a_{9}\)).
Given an arithmetic sequence where:
• First term: \(a_1 = 10\)
• The common difference: \(d = 7\)
Find the 9-th term (i.e. \(a_{9}\)).
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: \(a_n = a_1 + (n-1) \cdot d\)
We will use the formula: \(a_n = a_1 + (n-1) \cdot d\)
\(a_{9} = 10 + (9-1) \cdot 7\)
\(a_{9} = 10 + 8 \cdot 7\)
\(a_{9} = 10 + 56 = 66\)
\(a_{9} = 10 + 8 \cdot 7\)
\(a_{9} = 10 + 56 = 66\)
Answer: 66