Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
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.
The recursive rule adds 2 to each preceding term. Thus, a_2=11, a_3=13, and a_4=15. The value 13 is the third term, so it is a close but incorrect option. Exam tip: start from a_1 and count the term positions carefully.
The governing concept is a recursive rule: each term is obtained from the immediately preceding term. Starting with a₁ = 28, subtract 4 repeatedly: a₂ = 28 − 4 = 24, a₃ = 24 − 4 = 20, a₄ = 20 − 4 = 16, and a₅ = 16 − 4 = 12. Hence option C is correct. The same result can be obtained by recognizing an arithmetic progression with common difference d = −4: a₅ = a₁ + (5 − 1)d = 28 + 4(−4) = 12. Option A is obtained by too many subtractions, option B does not correspond to any term, and option D results from subtracting only 3 from the initial value.
By the recursive rule, each new term is twice the previous term. Thus, \(a_2=2\times6=12\), \(a_3=2\times12=24\), and \(a_4=2\times24=48\). Therefore, 48 is correct. The value 24 is \(a_3\), so it is a close but incorrect option. Exam tip: start from \(a_1\) and count the term positions carefully.
The recursive rule makes each new term half of the preceding term. Thus, \(a_2=96/2=48\), \(a_3=48/2=24\), and \(a_4=24/2=12\). Option 24 is \(a_3\), not \(a_4\). Exam tip: apply the recursive rule step by step from the given first term until the required term is reached.
In the recursive rule, add the current value of n at each step. Thus, a_2=4+1=5, a_3=5+2=7, a_4=7+3=10, and a_5=10+4=14. Therefore, 14 is correct. The value 16 would require adding 6 in the final step instead of 4, which does not follow the given rule. Exam tip: while finding successive terms, use n=1, 2, 3, ... in order.
The recursive rule subtracts the index at each step. Thus, \(a_2=22-1=21\), \(a_3=21-2=19\), and \(a_4=19-3=16\). Therefore, 16 is correct. Note that 19 is \(a_3\), not \(a_4\). Exam tip: to find \(a_4\), apply the rule successively for \(n=1,2,3\).
In the recursive rule, add \(3n\) to the current term at each step. \(a_2=5+3(1)=8\), \(a_3=8+3(2)=14\), and \(a_4=14+3(3)=23\). Therefore, the correct answer is 23. The option 21 would result from not applying the final increment \(3n\) correctly. Exam tip: to find \(a_4\), apply the rule successively for \(n=1,2,3\).
Using n=1, 2, and 3 in the recursive rule gives a_2=8+1=9, a_3=9+3=12, and a_4=12+5=17. Therefore, 17 is correct. The option 16 is incorrect because the successive odd numbers 1, 3, and 5 must be added. Exam tip: To find a_4 from a_1, apply the recursive rule three times.
By the recursive rule, each next term is 4 times the preceding term. Thus, \(a_2=4\times3=12\) and \(a_3=4\times12=48\). Therefore, 48 is correct. The value 24 would result from multiplying \(a_2\) by 2, so it does not follow the given rule. Exam tip: to find \(a_3\), apply the rule twice starting from \(a_1\).
The recursive rule means that each new term is found by multiplying the previous term by 2 and then adding 1. Thus, \(a_2=2\times3+1=7\), and \(a_3=2\times7+1=15\). Therefore, 15 is correct. A value such as 13 can result from applying the rule incorrectly. Exam tip: to find \(a_3\), calculate \(a_2\) first.
Apply the recursive rule step by step: \(a_2=2\times7-3=11\). Then \(a_3=2\times11-3=19\). Therefore, the correct answer is 19. The value 22 would result from forgetting to subtract 3 in the second step. Exam tip: for every new term, first multiply the previous term by 2 and then subtract 3.
The governing concept is a recursive sequence rule. The relation a_{n+1}=a_n+7 says that every new term is obtained by adding 7 to the immediately preceding term. Starting with a_1=2, calculate successively: a_2=2+7=9, a_3=9+7=16, a_4=16+7=23, and a_5=23+7=30. Therefore option B is correct. The same result follows from the arithmetic-sequence formula a_n=a_1+(n−1)d: a_5=2+4(7)=30. Option A is the fourth term, while options C and D add 7 too many times. The recurrence must be applied exactly four times after the first term to reach the fifth term.
The recursive rule makes each new term 6 less than the preceding term. Thus, the terms are 60, 54, 48, 42, 36, 30. Therefore, the sixth term, (a_6), is 30. The value 36 is the fifth term, so it is a close but incorrect option. Exam tip: from the first term to the sixth term, subtract 6 five times.
This is governed by a second-order recursive rule: every term equals the sum of the two immediately preceding terms. The initial values are a_1 = 4 and a_2 = 7. Therefore, a_3 = 7 + 4 = 11. Then a_4 = a_3 + a_2 = 11 + 7 = 18. Finally, a_5 = a_4 + a_3 = 18 + 11 = 29. Hence option D is correct. Option A is only a_4, option B does not arise from the stated sequence, and option C reflects an arithmetic or indexing mistake. It is important to use the two latest available terms at each step, not repeatedly add the original starting values.
Each new term is the sum of the two immediately preceding terms. Thus, \(a_3=8+3=11\), and \(a_4=11+8=19\). Therefore, the correct answer is 19. The value 16 would result from adding \(8+8\), but the recursive rule requires adding \(a_3\) and \(a_2\). Exam tip: find \(a_3\) first, then use it to calculate \(a_4\).
The recursive rule makes each new term one-third of the preceding term. Thus, \(a_2=\frac{45}{3}=15\) and \(a_3=\frac{15}{3}=5\). Option 15 is \(a_2\), not \(a_3\). Exam tip: to find \(a_3\) from \(a_1\), apply the rule twice.
Using the recursive rule, add 4 to each term: 5, 9, 13, 17, 21. Therefore, 21 is the 5th term. The 4th term is 17, so the fourth-term option is incorrect. Exam tip: list the terms with their positions in recursive-sequence questions.
The governing concept is locating a value in a recursively defined arithmetic sequence. The initial term is a₁ = 10, and each next term is obtained by adding 5 to the preceding term. Successive calculation gives a₁ = 10, a₂ = 15, a₃ = 20, a₄ = 25, and a₅ = 30. Therefore 30 is the fifth term, so option B is correct. The same result follows from the arithmetic-sequence formula aₙ = a₁ + (n−1)d, where d = 5. Setting the value equal to 30 gives 30 = 10 + 5(n−1), so 20 = 5(n−1), n−1 = 4, and n = 5. Option A is 25, the fourth term. Options C and D correspond to 35 and 40, which occur after 30 and therefore cannot be the requested position.
The governing concept is a recursive rule: it must identify the starting term and describe how each term is obtained from the immediately preceding term. The first term of the sequence is 12, so the initial condition must be a₁ = 12. Next, compare consecutive terms: 15 − 12 = 3, 18 − 15 = 3, and 21 − 18 = 3. The same difference is added at every stage, so the recurrence is aₙ₊₁ = aₙ + 3. Checking it gives a₂ = 12 + 3 = 15, a₃ = 15 + 3 = 18, and a₄ = 18 + 3 = 21. Thus option A is correct. Option B adds 4, option C starts with the wrong first term, and option D multiplies by 3, producing 12, 36, … rather than the given sequence.
A recursive rule must give the starting value and describe how to obtain each term from the preceding term. Here the first term is 36, and the differences are 31−36 = −5, 26−31 = −5, and 21−26 = −5. Therefore, subtracting 5 each time gives aₙ₊₁ = aₙ − 5. Option A has the wrong first term, B increases, and D multiplies.
A recursive rule tells us how to obtain each term from the term immediately before it. The first term here is 3. Comparing consecutive terms shows that 15 is obtained from 3 by multiplying by 5, 75 is obtained from 15 by multiplying by 5, and 375 is obtained from 75 by multiplying by 5. Therefore the same multiplication is repeated at every step.
In symbols, the starting value is written as \(a_1=3\), and the repeated operation is \(a_{n+1}=5a_n\). Thus option C gives both the correct first term and the correct rule. Option A uses a constant addition of 12, which fails because 15 to 75 is not an increase of 12. The ratio between successive terms is the useful clue.
A recursive rule must include the correct initial term as well as the operation used to produce the next term. The sequence begins with 72. Dividing 72 by 3 gives 24, and dividing 24 by 3 gives 8. Therefore each new term is one-third of the preceding term. The rule is not based on subtracting a fixed number, because the differences are 48 and 16, which are not equal.
The correct symbolic description is \(a_1=72\) and \(a_{n+1}=\frac{a_n}{3}\). This is option C. Option B uses the correct division operation but starts with 24, so it would produce 24, 8, and a later term instead of the stated sequence beginning at 72. Both the starting value and the recursive operation must match.
The governing concept is repeated use of a recursive relation. The condition a₁ = 20 gives the first term, and aₙ₊₁ = aₙ + 1 means that exactly 1 must be added whenever we move from one term to the next. To reach a₄ from a₁, there are three transitions, not four: a₂ = a₁ + 1 = 20 + 1 = 21; a₃ = a₂ + 1 = 21 + 1 = 22; and a₄ = a₃ + 1 = 22 + 1 = 23. Therefore option C is correct. The arithmetic-progression check gives a₄ = a₁ + (4 − 1)d = 20 + 3(1) = 23. Option A is the second term, option B is the third term, and option D results from adding 1 four times instead of the required three transitions.
By the recursive rule, each new term is 25 less than the previous term. Thus the terms are 100, 75, 50, 25. Therefore, a₄ = 25. The value 50 is the third term, so it is a close but incorrect option. Exam tip: to find a₄ from a₁, subtract 25 three times.
The recursive rule adds 6 to each preceding term. Thus the sequence is 4, 10, 16, 22, 28. Hence, \(a_2=10\) and \(a_5=28\), so \(a_5-a_2=28-10=18\). The value 24 would result from counting four common differences, but there are only three differences between \(a_2\) and \(a_5\). Exam tip: check the difference between the term indices when subtracting terms of an arithmetic sequence.
QUIZ COMPLETE
