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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.
Using the recursive rule, \(a_2=2\times6+4=16\), \(a_3=2\times16+4=36\), and \(a_4=2\times36+4=76\). Hence, \(a_4-a_2=76-16=60\), so option D is correct. \(56\) results from an incorrect calculation of the terms. Exam tip: List the required terms in order before finding their difference.
Given \(a_1=6\), the recursive rule gives \(a_2=2\times6+4=16\) and \(a_3=2\times16+4=36\). Therefore, \(a_3-a_2=36-16=20\). Note that 16 is only the second term, not the required difference. Exam tip: in recursive-sequence questions, calculate the required terms in order before finding their difference.
By the recursive rule, each new term is obtained by adding the previous two terms and 4. Thus, \(a_3=5+9+4=18\), \(a_4=9+18+4=31\), and \(a_5=18+31+4=53\). Therefore, the correct answer is 53. A value such as 49 can result from missing the constant 4 in a step. Exam tip: write each term in order and include the stated constant at every step.
The recursive rule requires adding the previous two terms and then adding 4 each time. First, \(a_3=9+5+4=18\). Then, \(a_4=18+9+4=31\). Therefore, the correct answer is \(31\). The value \(29\) would result from adding only 2 instead of 4 in the final step, so it does not follow the given rule. Exam tip: write each intermediate term in order until you reach the required term.
The rule adds 4 to every preceding term, so the difference between consecutive terms is constantly 4. Hence it is an arithmetic progression. A difference of \(-4\) would make terms decrease. Exam tip: identify an AP by checking for a constant difference.
In option A, the difference between consecutive terms is \(a_{n+1}-a_n=d\), which is constant; hence it is an arithmetic progression. In option C, the difference \(n\) changes. Exam tip: check for a constant difference in an AP.
Apply the recursive rule using the correct value of \(n\) at each step. For \(n=1\), \(a_2=4+7(1)-2=9\). Then, with \(n=2\), \(a_3=9+7(2)-2=21\). Hence, 21 is correct. A value such as 20 can result from using an incorrect value for \(7n-2\) in the second step. Exam tip: use \(n=1\) to find \(a_2\), and \(n=2\) to find \(a_3\).
Using the recursive rule with \(n=1\), \(a_2=150-(5\times1+4)=141\). Then, with \(n=2\), \(a_3=141-(5\times2+4)=141-14=127\). Therefore, \(127\) is correct. A close distractor such as \(124\) can result from calculating \(5n+4\) incorrectly in the second step. Exam tip: use the next value of \(n\) for each successive term.
(a_3=3\times2-2\times1+1=5) and (a_4=3\times5-2\times2+1=12). The correct option should include (12).
In an arithmetic progression, each term is obtained by adding a fixed common difference to the previous term. Since the difference is 3, \(a_n=a_{n-1}+3\). Exam tip: look for constant addition in a recursive AP rule.
The direct answer is option C, 20. A recursive rule tells us how to move from one term to the next. Starting with \(a_1=5\), we need four additions to reach \(a_5\): use \(n=1,2,3,4\). The additions are \(\frac{3(1)}2=\frac32\), \(\frac{3(2)}2=3\), \(\frac{3(3)}2=\frac92\), and \(\frac{3(4)}2=6\). Their total is \(\frac32+3+\frac92+6=15\), or \(\frac{3(1+2+3+4)}2=15\). Therefore \(a_5=5+15=20\). Option C is correct. Option A, 18, misses 2; option B, 19, misses 1; and option D, 21, adds 1 too much. The common mistake is to use only three increments, but moving from term 1 to term 5 requires four steps. Memory cue: for \(a_5\), add the changes for indices 1 through 4.
The direct answer is option B, 25. Begin with \(a_1=40\). The rule subtracts \(\frac{3n}{2}\), and reaching the fifth term requires n=1,2,3,4. The subtractions are \(\frac32,3,\frac92,6\). Their total is \(\frac32+3+\frac92+6=15\), also \(\frac{3(1+2+3+4)}2=15\). Thus \(a_5=40-15=25\). Option B is correct. Option A, 23, subtracts 17 instead of 15; option C, 27, subtracts only 13; and option D, 29, subtracts only 11. Notice that every change is subtracted, because the recurrence contains a minus sign. A frequent mistake is to add the changes or to stop after three changes. Exam cue: from the first term to the fifth term, use exactly four updates, with indices 1 to 4.
In the recursive rule, use the current index to find the next term. Thus, \(a_2=1\times2+3=5\), \(a_3=2\times5+3=13\), and \(a_4=3\times13+3=42\). Therefore, the correct answer is \(42\). An option such as \(39\) results from using the multiplier 3 or the added 3 incorrectly while finding \(a_4\). Exam tip: for \(a_{n+1}\), substitute \(n=1\), then \(n=2\), and then \(n=3\) in order.
Under \(a_{n+1}=r a_n\), the ratio \(a_{n+1}/a_n=r\) remains constant, so the sequence is geometric. In option B, the difference is constant, which defines an arithmetic progression. Exam tip: check whether a rule fixes a ratio or a difference.
The correct rule is \(a_1=3,\ a_{n+1}=2a_n+2\). It gives \(2\times3+2=8\), then \(2\times8+2=18\), and \(2\times18+2=38\), matching every listed term. Option B gives \(7\) as the second term, while option C has the wrong first term. Exam tip: test a recursive rule by generating at least three consecutive terms.
In option B, \(a_1=10\). Applying the rule gives \(2(10)-3=17\), \(2(17)-3=31\), and \(2(31)-3=59\), so it matches every given term. In option A, the next term would be 24, while option D has the wrong first term. Exam tip: when checking a recursive rule, verify the initial term and at least two subsequent terms.
The recursive rule subtracts 1², 2², 3², and so on. Thus, a₁=100, a₂=100−1=99, a₃=99−4=95, a₄=95−9=86, and a₅=86−16=70. Therefore, 70 is the fifth term. The fourth term is 86, so it is a close but incorrect option. Exam tip: in a recursive rule, use the value of n while moving from one term to the next.
Apply the recursive rule step by step: \(a_3=2+7+3=12\), \(a_4=7+12+4=23\), and \(a_5=12+23+5=40\). Therefore, the correct answer is 40. A value such as 38 can result from forgetting to add 5 while finding \(a_5\). Exam tip: for every new term, add both preceding terms and the index of the term being found.
First find the third term using the recursive rule: (a_3=7+2+3=12). Then, (a_4=a_3+a_2+4=12+7+4=23). Therefore, the correct answer is 23. The value 21 may result from forgetting to add 4 in the final step, but the rule requires adding the current index n. Exam tip: apply a recursive rule step by step until you reach the required term.
Given \(a_1=9\), first use \(n=1\) in the recursive rule: \(a_2=a_1+a_1+1^2=9+9+1=19\). Next, use \(n=2\): \(a_3=a_2+a_1+2^2=19+9+4=32\). Hence, the correct answer is \(32\). A choice such as \(34\) can result from an error in using the square term or the previous term. Exam tip: substitute the correct value of \(n\) at every step of a recurrence relation.
Given \(a_1=9\), substitute \(n=1\) in the recursive rule to find the second term: \(a_2=a_1+a_1+1^2=9+9+1=19\). Hence, 19 is correct. The value 17 would result from incorrectly omitting the \(n^2\) term. Exam tip: To find \(a_{n+1}\), first substitute the appropriate value of \(n\) into the recursive rule.
The recursive rule generates each new term from the preceding term. Thus, \(a_2=2\times6+1=13\), \(a_3=2\times13+2=28\), and \(a_4=2\times28+3=59\). Hence, the correct answer is 59. An answer such as 57 can result from incorrectly adding 1 instead of using \(n=3\) in the final step. Exam tip: write the value of \(n\) for each transition before substituting it into the rule.
The governing concept is a recursive sequence: each term is calculated from the preceding term using the current value of n. Begin with a₁ = 80. For n = 1, subtract 3(1)+2 = 5, so a₂ = 80 - 5 = 75. For n = 2, subtract 3(2)+2 = 8, giving a₃ = 75 - 8 = 67. For n = 3, subtract 3(3)+2 = 11, so a₄ = 67 - 11 = 56. Finally, for n = 4, subtract 3(4)+2 = 14, giving a₅ = 56 - 14 = 42. Thus option B is correct. The successive deductions are 5, 8, 11, and 14; each rises by 3. Skipping a transition or using the wrong index can produce the other choices.
Substituting \(n=1,2,3,4\) into the recursive rule gives successive increments of \(4,10,18,28\). Thus, \(a_2=1+4=5\), \(a_3=5+10=15\), \(a_4=15+18=33\), and \(a_5=33+28=61\). Therefore, 61 is correct. A value such as 63 would result from an incorrect addition at one of the steps. Exam tip: to find \(a_5\), apply the rule from \(n=1\) through \(n=4\).
In an arithmetic progression, the difference between consecutive terms is constant. The rule \(a_{n+1}=a_n+7\) adds exactly 7 at every step, so its common difference is 7. In \(a_n+n\), the difference changes with \(n\). Exam tip: compute \(a_{n+1}-a_n\); it must be a constant for an AP.
QUIZ COMPLETE
