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SubjectsMathematics
पुनरावर्ती नियम
In Class 9 Mathematics, the topic Recursive Rule in Sequences and Progressions explains how a sequence can be defined by giving one or more starting terms and a rule that uses earlier terms to find the next one. Students learn to read and write such rules, generate sequence terms step by step, recognize patterns, and check whether a rule correctly describes a sequence. The topic also connects recursive descriptions with familiar arithmetic and geometric progressions, helping students understand how terms change and how sequence patterns can be represented mathematically.
TOPIC PRACTICE
Up to 25 questions from this page. Select your focus, then start.
By the recursive rule, each new term is found by adding 3 to the previous term. Thus, \(a_2=2+3=5\), \(a_3=5+3=8\), and \(a_4=8+3=11\). Therefore, 11 is correct. The value 8 is the third term, not the fourth. Exam tip: in recursive-rule questions, list the terms successively until the required term.
The recursive rule subtracts 2 from each preceding term: \(a_1=10\), \(a_2=8\), \(a_3=6\), \(a_4=4\), and \(a_5=2\). Therefore, the correct answer is \(2\). \(4\) is the fourth term, not the fifth. Exam tip: write the term numbers while expanding a recursive sequence.
By the recursive rule, each new term is twice the preceding term. Thus, \(a_2=2\times3=6\), \(a_3=2\times6=12\), and \(a_4=2\times12=24\). Therefore, 24 is correct. Note that 12 is the value of \(a_3\), not \(a_4\). Exam tip: write the terms one by one until you reach the required term.
By the recursive rule, each new term is half of the preceding term. Thus, \(a_2=64/2=32\), \(a_3=32/2=16\), and \(a_4=16/2=8\). Therefore, the correct answer is 8. Option 16 is the third term, \(a_3\), not the fourth term. Exam tip: Start counting from \(a_1\) and apply the given rule at every step.
Each new term is obtained by adding the current index n to the preceding term. Thus, a_2=1+1=2, a_3=2+2=4, a_4=4+3=7, and a_5=7+4=11. Therefore, the correct answer is 11. The value 9 can result from using an incorrect index at a step. Exam tip: when finding a_{n+1}, use the current value of n, not n+1.
In the recursive rule, subtract the current value of n at each step. Thus, \(a_2=20-1=19\), \(a_3=19-2=17\), and \(a_4=17-3=14\). Therefore, 14 is correct. The value 16 would result from stopping after only two subtraction steps. Exam tip: to find \(a_4\), apply the rule successively for \(n=1,2,3\).
Apply the recursive rule successively with n=1, 2, and 3: a_2=4+2(1)=6, a_3=6+2(2)=10, and a_4=10+2(3)=16. Therefore, the correct answer is 16. The value 14 would result from incorrectly using n=2 in the final step; to find a_4, the last step uses n=3. Exam tip: starting from a_1, apply the rule three times to obtain a_4.
Apply the recursive rule successively for n=1, 2, and 3. We get a_2=5+(2×1−1)=6, a_3=6+(2×2−1)=9, and a_4=9+(2×3−1)=14. Therefore, the correct answer is 14. The value 16 would result from incorrectly using n=4 for the next step, which is not needed here. Exam tip: To find a_4 from a_1, apply the rule three times.
Each new term is 3 times the preceding term. Thus, \(a_2=3\times2=6\) and \(a_3=3\times6=18\). Therefore, the correct answer is 18. Option 6 is the second term, not the third. Exam tip: evaluate recursive sequences one term at a time.
The recursive rule says that each new term is twice the previous term plus 1. Thus, a_2=2(1)+1=3, a_3=2(3)+1=7, and a_4=2(7)+1=15. Therefore, the correct answer is 15. The value 7 is a close distractor because it is a_3, not a_4. Exam tip: start with the given first term and apply the rule one step at a time.
Given \(a_1=5\) and \(a_{n+1}=2a_n-1\), we get \(a_2=2\times5-1=9\) and then \(a_3=2\times9-1=17\). Therefore, 17 is correct. A value such as 19 results from using an incorrect previous term. Exam tip: in a recursive rule, calculate each term step by step from the preceding term.
The recursive rule adds 5 to each preceding term. Thus the terms are 2, 7, 12, 17, 22, 27, so the sixth term is 27. The value 22 is the fifth term, making it a close but incorrect distractor. Exam tip: for a constant-add recursive rule, keep adding the same number until the required term.
The recursive rule makes each new term \(4\) less than the preceding term. Thus the terms are \(30, 26, 22, 18, 14\). Therefore, \(a_5=14\). Note that \(16\) is the fourth term, \(a_4\), not the fifth. Exam tip: list the terms in order and count their indices carefully.
By the recursive rule, each new term is the sum of the two preceding terms. Thus, \(a_3=2+3=5\), \(a_4=3+5=8\), and \(a_5=5+8=13\). Therefore, the correct answer is \(13\). Note that \(8\) is \(a_4\), not \(a_5\). Exam tip: write each intermediate term in order until you reach the required term.
By the recursive rule, each new term is the sum of the two preceding terms. Thus, \(a_3=4+1=5\) and \(a_4=5+4=9\). Therefore, the correct answer is 9. A value such as 10 results from adding incorrect terms. Exam tip: find \(a_3\) first, then use it to calculate \(a_4\).
The recursive rule divides each term by 3 to obtain the next term. Thus, \(a_2=\frac{12}{3}=4\), and \(a_3=\frac{4}{3}\). Option 2 would result from dividing 12 by 3 only once, so it is \(a_2\), not \(a_3\). Exam tip: to find \(a_3\) from \(a_1\), apply the recursive rule twice.
The recursive rule increases each term by 2. Thus, the terms are 3, 5, 7, 9, 11. Therefore, 11 is the fifth term. The sixth term would be 13, so the sixth option is not correct. Exam tip: list the terms in order and count their positions.
The governing concept is a recursive rule with a constant increase of 4. Starting from a_1 = 4, the terms are 4, 8, 12, 16, and 20. Counting from the first position shows that 20 is a_5, the fifth term. Therefore option B is correct. The fourth term is only 16, while the sixth and seventh terms would be 24 and 28.
A recursive rule must contain two parts: an initial value and a relationship that generates each next term from the preceding term. The first term here is 5. The successive differences are 8−5=3, 11−8=3, and 14−11=3, so the same operation is repeated each time: add 3 to the previous term. Consequently, the correct rule is a_1 = 5 and a_(n+1) = a_n + 3, which is option A. Option B uses a difference of 4 and would produce 5, 9, 13, ... . Option C starts with the second term rather than the first, and option D doubles each term, producing a completely different sequence. Both the starting value and recurrence must agree with the displayed sequence.
A recursive rule describes a sequence by giving its starting term and a rule for obtaining each next term from the preceding term. In the sequence \(20,15,10,5,\ldots\), the first term is clearly 20. Comparing consecutive terms shows that each term is 5 less than the term immediately before it.
Starting with \(a_1=20\), the rule \(a_{n+1}=a_n-5\) produces \(a_2=15\), then \(a_3=10\), and then \(a_4=5\), exactly matching the sequence. Adding 5 would make the sequence increase, and starting with 15 would shift every position. Multiplication by 5 also does not fit. Therefore option C is correct.
A recursive rule must state the initial term and explain how to obtain each following term from the preceding one. The sequence starts with a_1=2. Each next value is twice the previous value: 4=2×2, 8=2×4, and 16=2×8. Therefore the recursive rule is a_1=2 and a_(n+1)=2a_n, which is option B. Option A adds 2 and would produce 2, 4, 6, 8 instead. Option C has the correct multiplier but starts with 4, so it describes a different indexed sequence. Option D subtracts 2 and produces decreasing terms. The constant multiplier also shows the related geometric pattern.
A recursive rule requires both the correct initial term and the correct transition from one term to the next. The sequence begins with a_1=81. Dividing successively by 3 gives 81÷3=27, 27÷3=9, and 9÷3=3. Thus the rule is a_1=81 and a_(n+1)=a_n÷3, which is option C. Option A multiplies by 3 and would produce 243 after 81. Option B uses division by 3 but starts at 27, so it does not match the given first term or indexing. Option D subtracts 3 and would produce 78 after 81. The repeated quotient also identifies the sequence as geometric with common ratio 1/3.
The governing concept is a recursively defined sequence: each new term is obtained from the preceding term by applying the stated rule. Start with a₁ = 7 and add 1 repeatedly until the sixth position is reached. Thus a₂ = 7 + 1 = 8, a₃ = 9, a₄ = 10, a₅ = 11, and a₆ = 12. Therefore option C is correct. Equivalently, this is an arithmetic sequence with first term 7 and common difference 1, so aₙ = 7 + (n − 1)×1; putting n = 6 gives 7 + 5 = 12. Option A stops at a₄, option B is a₅, and option D applies the rule one extra time to obtain a₇.
The recursive rule makes each new term 10 less than the previous term. Thus, a_2=40, a_3=30, and a_4=20. Therefore, 20 is correct. The value 30 is the third term, not the fourth. Exam tip: for a recursive sequence, list the terms step by step up to the required term.
The recursive rule increases each term by 2. Thus the terms are 6, 8, 10, 12, 14. Hence \(a_3=10\) and \(a_5=14\), so \(a_5-a_3=14-10=4\). The value 2 is the increase for one step only; there are two steps from \(a_3\) to \(a_5\). Exam tip: when subtracting two terms, count the number of steps between their indices.
QUIZ COMPLETE
