Sequences

The next difference is (15), so the sixth term is (47+15=62). Exam tip: write changing differences separately.

This is a triangular sequence with (T_n=\frac{n(n+1)}{2}). Since (\frac{9\times10}{2}=45), it is the (9)th term.

This is a geometric sequence with first term 2 and common ratio 2. Therefore, \(a_8=2\times 2^{7}=2^{8}=256\). The value 128 is the seventh term, so it is a close but incorrect option. Exam tip: write terms of a doubling sequence as powers of 2.

Given \(a_n=n^2+3\) and \(a_n=84\), we get \(n^2+3=84\). Thus, \(n^2=81\), so \(n=9\). Therefore, 84 is the 9th term of the sequence. The 8th term is \(8^2+3=67\), so it is not correct. Exam tip: to find a term number, substitute the given term value in the formula for \(a_n\) and solve for \(n\).

The differences between consecutive terms are \(13-6=7\), \(24-13=11\), \(39-24=15\), and \(58-39=19\). These differences increase by \(4\) each time, so the next difference is \(23\). Hence, the next term is \(58+23=81\). Choosing \(80\) would give a difference of \(22\), which does not follow the pattern. Exam tip: For such sequences, first list consecutive differences and look for their pattern.

The nth term is n^3. Therefore, the 6th term is 6^3 = 216 and the 5th term is 5^3 = 125. Hence, the difference is 216 - 125 = 91. Note that 81 is 4^3, not the required difference. Exam tip: evaluate both specified terms first, then subtract the smaller term from the larger one.

Odd positions are (10,11,12,\ldots), so the next odd-position term is (13). Exam tip: separate positions in alternating sequences.

Odd positions are (2,4,6,\ldots), so the next odd-position term is (8). Exam tip: split the two patterns in alternating sequences.

The general term is given as (a_n=n^2+2). Substituting n=9, we get (a_9=9^2+2=81+2=83). The value 81 is only (9^2); the additional 2 must also be included. Exam tip: when a general term is given, substitute the required term number directly for n.

This is an arithmetic sequence with first term 12 and common difference 5. Its nth term is \(a_n=12+5(n-1)\). Using \(a_n<60\), we get \(12+5(n-1)<60\), so \(n<10.6\). Thus, the possible whole-number values of \(n\) are 1 through 10, giving 10 terms less than 60. The 11th term is 62, so it is not included. Exam tip: for “less than,” use the strict inequality \(<\), not \(\leq\).

In option A, the consecutive differences are 5−2=3 and 8−5=3, so the difference is constant. However, the ratios 5/2 and 8/5 differ. In exams, check differences and ratios separately.

This is a geometric sequence with first term \(a=5\) and common ratio \(r=2\). Its \(n\)th term is \(a_n=5\times2^{n-1}\). From \(5\times2^{n-1}=320\), we get \(2^{n-1}=64=2^6\). Therefore, \(n-1=6\) and \(n=7\). Hence, 320 is the 7th term. The 6th term is \(160\), so it is not correct. Exam tip: use \(a_n=ar^{n-1}\) to find a term's position in a geometric sequence.

The sixth term is (16+26=42) and the seventh is (26+42=68). Exam tip: count positions carefully in recursive sequences.

The governing method is to extend the given sequence of successive differences. The differences are 3, 6, 9, and 12, which increase by 3 each time; hence the next difference is \(12+3=15\). The sixth term is obtained by adding this next difference to the fifth term: \(31+15=46\). Therefore option C is correct. One might choose 43 by adding 12 again, but that ignores the increase in the differences. Values 45 and 49 likewise do not result from the established difference sequence.

The sequence of differences is itself arithmetic: \(-5,-8,-11,-14\), with a common difference of \(-3\). Therefore the next difference is \(-14-3=-17\). Apply it to the last known term, 12: \(12+(-17)=12-17=-5\). Thus option C is correct. Option A would use a difference of -12, while options B and D do not follow the continuing decrease of 3 in each successive difference. Keeping the negative sign attached to the difference is essential; treating 17 as positive would incorrectly produce 29.

Substituting (n=1,2,3) in (n^2+2n-1) gives (2,7,14). Exam tip: test options using the first three terms.

The terms are (1^2-1,2^2-1,3^2-1,\ldots). Exam tip: quickly recognize sequences near perfect squares.

The consecutive differences are 4, 6, 8, and 10, increasing by 2 each time, so the sequence is quadratic in nature. Substituting \(n=1,2,3,4,5\) in \(a_n=n^2+n+1\) gives 3, 7, 13, 21, and 31 respectively. Option C gives 7 when \(n=2\), but it gives 1 rather than 3 when \(n=1\). Exam tip: verify a proposed general term using at least the first three or four terms.

The differences between consecutive terms are 18−11=7, 27−18=9, 38−27=11, and 51−38=13. Each difference increases by 2, so the next difference is 15. Hence, the next term is 51+15=66. Choosing 64 would mean adding 13 again, which does not follow the increasing-difference pattern. Exam tip: For such sequences, first list the consecutive differences and check their pattern.

This question has inconsistent options because (n^3+n) gives (10) at (n=2), not (12). Exam tip: verify every option before accepting a pattern.