Q1: What is the next term in the arithmetic sequence 7, 12, 17, 22, ...? (a) 25 (b) 27 (c) 29 (d) 32
Solution:
Ans: (b) Explanation: This is an arithmetic sequence with a common difference of 5 (12 - 7 = 5). To find the next term, add 5 to 22: 22 + 5 = 27.
Q2: Which formula represents the nth term of an arithmetic sequence? (a) \(a_n = a_1 \cdot r^{n-1}\) (b) \(a_n = a_1 + (n-1)d\) (c) \(a_n = a_1 + nd\) (d) \(a_n = n \cdot d\)
Solution:
Ans: (b) Explanation: The explicit formula for an arithmetic sequence is \(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number. Option (a) is for geometric sequences, option (c) has an incorrect coefficient, and option (d) is missing the first term.
Q3: What is the common ratio of the geometric sequence 3, 6, 12, 24, ...? (a) 3 (b) 2 (c) 6 (d) 4
Solution:
Ans: (b) Explanation: In a geometric sequence, the common ratio is found by dividing any term by the previous term. \(\frac{6}{3} = 2\), \(\frac{12}{6} = 2\), and \(\frac{24}{12} = 2\). The common ratio is 2.
Q4: If the 5th term of an arithmetic sequence is 23 and the common difference is 4, what is the first term? (a) 3 (b) 5 (c) 7 (d) 11
Solution:
Ans: (c) Explanation: Using the formula \(a_n = a_1 + (n-1)d\), we have: \(23 = a_1 + (5-1)(4)\) \(23 = a_1 + 16\) \(a_1 = 23 - 16 = 7\) The first term is 7.
Ans: (b) Explanation: A geometric sequence has a constant ratio between consecutive terms. In option (b), \(\frac{10}{5} = 2\), \(\frac{20}{10} = 2\), \(\frac{40}{20} = 2\), and \(\frac{80}{40} = 2\). Options (a) and (d) are arithmetic sequences with constant differences, and option (c) is a triangular number sequence.
Q6: What is the 8th term of the geometric sequence where \(a_1 = 2\) and \(r = 3\)? (a) 4374 (b) 2187 (c) 6561 (d) 1458
Solution:
Ans: (a) Explanation: Using the geometric sequence formula \(a_n = a_1 \cdot r^{n-1}\): \(a_8 = 2 \cdot 3^{8-1}\) \(a_8 = 2 \cdot 3^7\) \(a_8 = 2 \cdot 2187 = 4374\)
Q7: An arithmetic sequence has a first term of 15 and a 10th term of 51. What is the common difference? (a) 3 (b) 4 (c) 5 (d) 6
Solution:
Ans: (b) Explanation: Using \(a_n = a_1 + (n-1)d\): \(51 = 15 + (10-1)d\) \(51 = 15 + 9d\) \(36 = 9d\) \(d = 4\) The common difference is 4.
Q8: Which term of the arithmetic sequence 5, 9, 13, 17, ... is equal to 73? (a) 16th term (b) 17th term (c) 18th term (d) 19th term
Solution:
Ans: (c) Explanation: First, identify that \(a_1 = 5\) and \(d = 4\). Use \(a_n = a_1 + (n-1)d\): \(73 = 5 + (n-1)(4)\) \(73 = 5 + 4n - 4\) \(73 = 1 + 4n\) \(72 = 4n\) \(n = 18\) The number 73 is the 18th term.
Section B: Fill in the Blanks
Q9:In an arithmetic sequence, the difference between consecutive terms is called the __________.
Solution:
Ans: common difference Explanation: The common difference is the constant value added to each term to get the next term in an arithmetic sequence.
Q10:The formula for the nth term of a geometric sequence is \(a_n = a_1 \cdot\) __________.
Solution:
Ans: \(r^{n-1}\) Explanation: In a geometric sequence, the nth term is found using the formula \(a_n = a_1 \cdot r^{n-1}\), where \(r\) is the common ratio.
Q11:If the first term of an arithmetic sequence is 8 and the common difference is -3, the second term is __________.
Solution:
Ans: 5 Explanation: The second term is found by adding the common difference to the first term: \(8 + (-3) = 5\).
Q12:A sequence in which each term after the first is obtained by multiplying the previous term by a constant is called a __________ sequence.
Solution:
Ans: geometric Explanation: A geometric sequence is characterized by a constant multiplier called the common ratio between consecutive terms.
Q13:In the geometric sequence 4, 12, 36, 108, ..., the common ratio is __________.
Solution:
Ans: 3 Explanation: The common ratio is found by dividing any term by the previous term: \(\frac{12}{4} = 3\).
Q14:The explicit formula for an arithmetic sequence requires knowing the first term and the __________.
Solution:
Ans: common difference Explanation: The explicit formula \(a_n = a_1 + (n-1)d\) requires both the first term \(a_1\) and the common difference \(d\).
Section C: Word Problems
Q15:A theater has 20 seats in the first row. Each row after the first has 3 more seats than the row before it. How many seats are in the 12th row?
Solution:
Ans: This is an arithmetic sequence problem where \(a_1 = 20\) and \(d = 3\). Using the formula \(a_n = a_1 + (n-1)d\): \(a_{12} = 20 + (12-1)(3)\) \(a_{12} = 20 + 11 \times 3\) \(a_{12} = 20 + 33\) \(a_{12} = 53\) Final Answer: 53 seats
Q16:A savings account starts with $500. Each month, the account balance increases by $75. What will be the account balance after 8 months?
Solution:
Ans: This forms an arithmetic sequence where \(a_1 = 500\) and \(d = 75\). We need the balance after 8 months, which is the 9th term (initial amount plus 8 increases): \(a_9 = 500 + (9-1)(75)\) \(a_9 = 500 + 8 \times 75\) \(a_9 = 500 + 600\) \(a_9 = 1100\) Final Answer: $1100
Q17:A ball is dropped from a height of 100 feet. After each bounce, it reaches 60% of its previous height. What height does the ball reach after the 4th bounce?
Solution:
Ans: This is a geometric sequence where \(a_1 = 100\) and \(r = 0.6\). After the 4th bounce is the 5th term in the sequence: \(a_5 = 100 \times (0.6)^{5-1}\) \(a_5 = 100 \times (0.6)^4\) \(a_5 = 100 \times 0.1296\) \(a_5 = 12.96\) Final Answer: 12.96 feet
Q18:The population of a town is growing according to a geometric sequence. If the current population is 8,000 and it doubles every 5 years, what will be the population after 15 years?
Solution:
Ans: Since the population doubles every 5 years, after 15 years there are 3 doublings. This is a geometric sequence where \(a_1 = 8000\) and \(r = 2\). After 15 years (which is the 4th term, counting the initial as the 1st): \(a_4 = 8000 \times 2^{4-1}\) \(a_4 = 8000 \times 2^3\) \(a_4 = 8000 \times 8\) \(a_4 = 64000\) Final Answer: 64,000 people
Q19:A ladder has rungs that are placed at regular intervals. The bottom rung is 6 inches from the ground, and each subsequent rung is 10 inches higher than the previous one. How high from the ground is the 15th rung?
Solution:
Ans: This is an arithmetic sequence where \(a_1 = 6\) and \(d = 10\). Using \(a_n = a_1 + (n-1)d\): \(a_{15} = 6 + (15-1)(10)\) \(a_{15} = 6 + 14 \times 10\) \(a_{15} = 6 + 140\) \(a_{15} = 146\) Final Answer: 146 inches
Q20:A scientist is observing bacteria growth in a petri dish. The bacteria count starts at 50 and triples every hour. How many bacteria will there be after 5 hours?
Solution:
Ans: This is a geometric sequence where \(a_1 = 50\) and \(r = 3\). After 5 hours, we need the 6th term (initial count plus 5 growth periods): \(a_6 = 50 \times 3^{6-1}\) \(a_6 = 50 \times 3^5\) \(a_6 = 50 \times 243\) \(a_6 = 12150\) Final Answer: 12,150 bacteria
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