Sequences

Each term in the sequence decreases by 7. Therefore, the first 5 terms are 72, 65, 58, 51, and 44. Their sum is 72+65+58+51+44=290, so option C is correct. A value such as 285 can result from an addition error. Exam tip: Write all the required terms in order before adding them.

This is an arithmetic progression with first term \(a=4\) and common difference \(d=6\). Its \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(4+(n-1)\times6=58\) gives \(n-1=9\), so \(n=10\). Therefore, 58 is the 10th term. The 9th term is \(52\), so it is not correct. Exam tip: To find the position of a given term in an AP, substitute that term for \(a_n\) in \(a_n=a+(n-1)d\).

This is an arithmetic sequence with first term 7 and common difference 5. Therefore, the sixth term is 32 and the seventh term is 37. Hence, \(a_6+a_7=32+37=69\). An option such as 64 results from using an incorrect term position or common difference. In exams, write the required terms separately before adding them.

This is an arithmetic sequence with common difference 7. Thus, \(a_4=39\) and \(a_7=18+6\times 7=60\). Therefore, \(a_7-a_4=60-39=21\). Option 18 may result from counting the gap between the terms incorrectly. Exam tip: In an arithmetic sequence, use \(a_n-a_m=(n-m)d\) to find the difference directly.

This is an arithmetic progression with common difference \(5\). Here, \(a_2=17\) and \(a_6=12+(6-1)\times5=37\). Therefore, \(a_6-a_2=37-17=20\). Choosing \(25\) incorrectly counts five intervals instead of the four intervals between the 2nd and 6th terms. Exam tip: use \(a_n-a_m=(n-m)d\) to find the difference between two terms.

In an arithmetic progression, each term is obtained by adding the same fixed number, called the common difference, to the preceding term. For example, in 5, 9, 13, 4 is added each time. A constant difference between consecutive terms is an equivalent statement, so the original options A and C would both have been correct; C is retained as the definitional form to ensure one correct answer. A constant ratio identifies a geometric progression. Exam tip: compare consecutive differences.

This is an arithmetic sequence with first term 96 and common difference \(-9\). To reach the seventh term, 9 is subtracted 6 times: \(a_7=96-6\times9=96-54=42\). Therefore, 42 is correct. The value 48 results from subtracting 9 only 5 times, which gives the sixth term. Exam tip: From the first term to the \(n\)th term, there are \(n-1\) changes.

The recursive rule is to multiply the previous term by 3 and then add 1. Thus, \(121\times 3+1=363+1=364\). Therefore, 364 is the correct next term. Option 366 would result from adding 3 after tripling 121, but the rule requires adding only 1. Exam tip: In recursive-sequence questions, apply the stated operation to the last given term in the same order.

The rule is to divide the previous term by 2 and then subtract 1. Therefore, the next term is \(10.25\div 2-1=5.125-1=4.125\). The value 5.125 is obtained only after division; the subtraction of 1 is still required. Exam tip: In recursive sequences, apply the operations in the stated order.

Look at the differences between consecutive terms: \(11-6=5\), \(21-11=10\), \(36-21=15\), and \(56-36=20\). The difference increases by \(5\) each time, so the next difference is \(25\). Hence, the next term is \(56+25=81\). Choosing 84 would give a difference of \(28\), which does not follow the pattern. Exam tip: for such sequences, first list the consecutive differences to identify the pattern.

The differences between consecutive terms are \(-10,-20,-30,-40\). Thus, the number being subtracted increases by 10 each time. The next difference is \(-50\), so the next term is \(100-50=50\). Getting 60 would require subtracting 40, but the next subtraction should be 50. Exam tip: Write consecutive differences first to identify a sequence pattern.

The important idea is to compare each term with the term immediately before it. Here, the sequence does not add the same number each time; instead, it uses changing multiplication factors. The factors follow a simple pattern: first multiply by 2, then by 3, and then by 4. Continuing this pattern means that the next factor should be 5. This gives the next term as 360, which corresponds to option D. A choice such as 288 would result from using 4 again, so it does not continue the stated pattern.

The calculation can be checked step by step: 3 times 2 equals 6, 6 times 3 equals 18, and 18 times 4 equals 72. Therefore, the next operation is 72 times 5. Using place-value multiplication, 70 times 5 is 350 and 2 times 5 is 10, giving 360 altogether. Thus the supplied answer D is consistent with the sequence and its explanation. The rule is inferred from the visible multiplier pattern, as is usual in this type of school sequence question.

The third term of the first sequence is 100, and that of the second sequence is 150. Therefore, their ratio is 100:150 = 2:3. The ratio 3:2 would represent the reverse order, from the second term to the first. In exams, keep the order stated in the question while writing a ratio.

This is an arithmetic sequence with first term 9 and common difference 6. Its first 5 terms are 9, 15, 21, 27, and 33. Their sum is 9+15+21+27+33=105, so the correct answer is 5. The sum of the first 4 terms is only 72. Exam tip: When the number of terms is small, add consecutive terms to verify the required sum quickly.

Each term in this sequence is obtained by adding 8 to the previous term: 8, 16, 24, 32, 40, 48, 56. Therefore, the term immediately before 56 is 48. Option 52 is not correct because it is not a term of this sequence. Exam tip: To find a previous term, first identify the common difference between consecutive terms.

In this sequence, each term is half of the preceding term: 192, 96, 48, 24, 12, 6. Therefore, the term immediately before 6 is 12. Option 18 is not correct because halving 24 gives 12, not 18. Exam tip: To identify a sequence rule, check the ratio of consecutive terms.

This is an arithmetic sequence because 5 is added to each term. Here, the first term is \(a=14\) and the common difference is \(d=5\). Therefore, \(a_n=a+(n-1)d=14+(n-1)\times5=5n+9\). In \(5n+14\), putting \(n=1\) gives 19, so it does not produce the first term. Exam tip: identify the first term and common difference, then use \(a_n=a+(n-1)d\).

This is an arithmetic progression with first term \(a_1=84\) and common difference \(d=-8\). Hence, \(a_n=a_1+(n-1)d=84-8(n-1)=92-8n\). In option B, putting \(n=1\) gives 76, so it does not match the first term. Exam tip: verify an nth-term rule by substituting \(n=1\) and checking the first term.

Each new term is obtained by adding the two preceding terms. Thus, a_3=6+11=17, a_4=11+17=28, and a_5=17+28=45. Hence, 45 is correct. An answer such as 39 can result from adding the terms incorrectly. Exam tip: In a recursive sequence, write each pair of previous terms before calculating the next term.

Each term is 4 times the preceding term. Therefore, the general term is \(a_n=5\times4^{n-1}\). From \(5\times4^{n-1}=1280\), we get \(4^{n-1}=256=4^4\). Hence, \(n-1=4\) and \(n=5\). For \(n=4\), the term is 320, not 1280. Exam tip: when consecutive terms have a common multiplier, treat the sequence as a geometric progression and use its general term.

The differences between consecutive terms are 3, 5, 7, and 9. These are consecutive odd numbers, so the next difference is 11. Therefore, the next term is \(34+11=45\). Getting 43 would require repeating the difference 9, which does not follow the increasing-difference pattern. Exam tip: For such sequences, first write the differences between consecutive terms and identify their pattern.

The differences between consecutive terms are \(18-12=6\), \(27-18=9\), and \(39-27=12\). Since these differences increase by 3 each time, the next difference is \(15\). Therefore, the next term is \(39+15=54\). Choosing 51 would give a difference of only 12, so it does not continue the pattern. Exam tip: Write the consecutive differences first, then identify their pattern.

Given \(a_n=2n^2+n\), \(a_5=2(5)^2+5=50+5=55\) and \(a_3=2(3)^2+3=18+3=21\). Hence, \(a_5-a_3=55-21=34\). Option C, 32, can result from evaluating a term incorrectly. In exams, substitute the value of \(n\) separately in each term before finding their difference.

The governing concept is the successive-difference method for identifying a sequence pattern. Subtract consecutive terms: 31−20=11, 46−31=15, 65−46=19, and 88−65=23. These first differences are 11, 15, 19, 23, which form an arithmetic progression with common difference 4. The next first difference is therefore 23+4=27. Adding it to the last displayed term gives 88+27=115. Hence option C is correct. The alternatives would require next differences of 23, 25, or 29, respectively, and none continues the observed first-difference pattern. The second differences are all 4, giving an additional check that 115 is the intended next term.

The governing idea is to examine the sequence of successive differences. Subtract consecutive terms: \(4-1=3\), \(10-4=6\), \(19-10=9\), and \(31-19=12\). These differences themselves increase by 3, so the next difference is \(12+3=15\). To obtain the sixth term, add this next difference to the fifth term: \(31+15=46\). Hence option C is correct. Option A would incorrectly add 12 again, ignoring the established increase in the differences. Options B and D also fail to produce the next term from the stated pattern. The important point is that the original terms are not in an arithmetic progression; their differences form the simpler pattern.