Sequences

Here, a_n=6n^2. Substituting n=1,2,3,4 gives 6(1)^2=6, 6(2)^2=24, 6(3)^2=54, and 6(4)^2=96. Therefore, the correct sequence is (6, 24, 54, 96). Option C gives only the values of n^2; it does not multiply them by 6. Exam tip: For a sequence rule, substitute the first few natural-number values of n to obtain its terms.

Here, \(a_n=2^n+n\), and the first term is obtained by taking \(n=1\). Thus, \(a_1=2^1+1=3\), \(a_2=2^2+2=6\), \(a_3=2^3+3=11\), and \(a_4=2^4+4=20\). Therefore, the correct sequence is \((3, 6, 11, 20)\). Option C is close, but its third term is incorrect because \(2^3+3=11\), not 12. Exam tip: evaluate the power first and then add \(n\).

Both sequences have common difference (6), so the term difference always remains (3). The (10)th terms also differ by (3).

Both sequences have difference (7), and their first terms differ by (9). So their (8)th terms also differ by (9).

This is an arithmetic sequence with first term 7 and common difference 7. Its first 8 terms are 7, 14, 21, 28, 35, 42, 49, and 56, whose sum is 252. Therefore, option C is correct. A value such as 246 can result from an error while adding a term. Exam tip: You can also use \(S_n=\frac{n}{2}[2a+(n-1)d]\) for an arithmetic sequence.

This is an arithmetic progression with first term 96 and common difference \(-8\). Its first 6 terms are 96, 88, 80, 72, 64, and 56, whose sum is 456. A value such as 464 results from an error while adding the decreasing terms. Exam tip: Use \(S_n=\frac{n}{2}[2a+(n-1)d]\) to find the sum of an arithmetic progression quickly.

This is an arithmetic progression with first term \(a=6\) and common difference \(d=13-6=7\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(76=6+(n-1)\times 7\), so \(70=7(n-1)\) and hence \(n=11\). Therefore, 76 is the 11th term. The 10th term is \(69\), so it is not correct. Exam tip: To find the position of a term in an AP, first identify \(a\) and \(d\), then use the \(n\)th-term formula.

This is an arithmetic sequence with first term 9 and common difference 5. Thus, the seventh term is 39 and the eighth term is 44. Therefore, \(a_7+a_8=39+44=83\), so option A is correct. A value such as 85 can result from using an incorrect term number or common difference. Exam tip: use \(a_n=a+(n-1)d\) to find any term of an arithmetic sequence.

This is an arithmetic progression with common difference 9. The third term is 38 and the eighth term is 83. Therefore, \(a_8-a_3=83-38=45\). Option 40 is incorrect because there are five gaps between the 3rd and 8th terms, each of size 9. Exam tip: For an AP, use \(a_m-a_n=(m-n)d\) to find the difference between two terms quickly.

In an arithmetic progression, consecutive differences are equal. Thus, \(b-a=c-b\), which rearranges to \(2b=a+c\). The relation \(b^2=ac\) is associated with three consecutive terms of a geometric progression. Exam tip: double the middle term.

This is an arithmetic sequence with first term 14 and common difference 8. To reach the 11th term, 8 is added 10 times: \(a_{11}=14+(11-1)\times 8=14+80=94\). Hence, 94 is correct. Getting 98 would mean adding 8 eleven times, which is incorrect for the 11th term. Exam tip: use \(a_n=a_1+(n-1)d\) for the nth term.

In option A, the successive differences are 3-7=-4, -1-3=-4 and -5-(-1)=-4. Since the difference is constant, it is an arithmetic progression. Negative terms do not prevent a sequence from being an AP. Exam tip: compare consecutive differences.

The recursive rule is to multiply the previous term by 3 and then add 2. Therefore, the term after 107 is \(3\times107+2=321+2=323\). Option 321 is only \(3\times107\); the required addition of 2 has not been made. Exam tip: Apply every step of the stated rule to the last given term in order.

The rule is to divide the preceding term by 2 and then subtract 1. Therefore, \(14.25\div2-1=7.125-1=6.125\). Hence, 6.125 is the correct next term. 7.125 is obtained only after division; the subtraction of 1 is still required. Exam tip: Apply the operations in the stated order—divide first, then subtract.

Look at the differences between consecutive terms: \(14-8=6\), \(26-14=12\), \(44-26=18\), and \(68-44=24\). The differences increase by \(6\) each time, so the next difference is \(30\). Therefore, the next term is \(68+30=98\). Choosing 96 would give a difference of only 28, which does not follow the pattern. Exam tip: For such sequences, first list the consecutive differences and check their pattern.

The successive subtractions are 12, 24, 36, and 48. These are increasing by 12 each time, so the next subtraction must be 60. Therefore, the next term is \(180-60=120\). Option 132 would result from subtracting 48 again, which repeats the previous subtraction. In exams, list the consecutive differences first to identify the pattern quickly.

This sequence is formed by multiplying consecutive terms by increasing whole numbers. The first transition uses a factor of 2, the second uses 3, and the third uses 4. The natural continuation is to use 5 next. Therefore, the missing term must be 96 times 5, which equals 480. This is option B. The sequence is not using a fixed multiplier, so treating it as a geometric sequence with one common ratio would be inappropriate.

Check every transition: 4 times 2 gives 8, 8 times 3 gives 24, and 24 times 4 gives 96. The next operation is consequently 96 times 5. Since 100 times 5 is 500 and 4 times 5 is 20, subtracting gives 500 minus 20, or 480. Hence option B follows directly from the multiplier pattern. The supplied answer and its short explanation correctly identify both the rule and the calculation. No ambiguity affects the intended school-level interpretation of this sequence.

The third term of the first sequence is \(150\), and that of the second sequence is \(200\). Hence, their ratio is \(150:200\). Dividing both terms by \(50\) gives \(3:4\). \(4:3\) would reverse the required order of the ratio. Exam tip: write ratios in the same first-to-second order as given in the question.

This is an arithmetic progression with first term \(a=10\) and common difference \(d=8\). The sum of \(n\) terms is \(S_n=\frac{n}{2}[2a+(n-1)d]\). So, \(130=\frac{n}{2}[20+8(n-1)]\), which gives \(n=5\). Indeed, the first five terms \(10,18,26,34,42\) add up to \(130\). The sum of six terms would be \(180\), so it is not correct. Exam tip: when the number of terms is unknown, substitute the given sum in the AP sum formula and solve for \(n\).

Each term in this sequence is obtained by adding 9 to the previous term: 12, 21, 30, 39, 48, 57, 66, 75. Therefore, the term immediately before 75 is 66. Option 69 is not correct because adding 9 to 66 gives 75, whereas adding 9 to 69 does not. Exam tip: identify the common difference and subtract it to find the previous term.