Sequences

Given \(a_n=2^n-n\), substitute \(n=8\): \(a_8=2^8-8=256-8=248\). Therefore, 248 is the correct option. A value such as 240 may result from evaluating the power incorrectly or subtracting at the wrong step. Exam tip: calculate \(2^8=256\) first, then subtract 8.

(a_{11}=275) and (a_{12}=324), so (11) terms are less than (300). Exam tip: check the two terms near the boundary.

The difference between the second and third terms is 15, so x = 12 + 15 = 27. Checking it, 50 - 27 = 23, which matches the next given difference. Hence, 27 is correct; if x were 30, the next difference would be 20, not 23. Exam tip: after finding a missing term, verify it using the differences on both sides.

The rule is: next term = twice the previous term + 2. Hence, the term after 62 is \(2\times 62+2=126\). Although 124 is twice 62, it does not include the required addition of 2. Exam tip: in a recursive sequence, apply the stated rule directly to the last given term.

The successive terms are multiplied by 2, 3, 4, and 5: \(2\times2=4\), \(4\times3=12\), \(12\times4=48\), and \(48\times5=240\). Therefore, the next multiplier is 6, so the next term is \(240\times6=1440\). Option 1200 would repeat multiplication by 5, which does not follow the increasing multiplier pattern. Exam tip: divide consecutive terms to identify a changing-multiplier pattern.

In B, \(a_{n+1}-a_n=[3(n+1)-2]-(3n-2)=3\) for every \(n\), so it is an AP. For \(n^2+1\), the difference changes. Exam tip: compare consecutive differences.

For \(a_n=n^2+1\), the terms are 2, 5, 10, 17. Its first differences are 3, 5, 7, while the second differences are 2, 2. A linear sequence has constant first differences instead. Exam tip: constant second differences indicate a quadratic sequence.

The direct answer is option B: the first term less than \(\frac{1}{50}\) is the 17th term. The sequence is \(\frac13,\frac16,\frac19,\ldots\), so its nth term is \(a_n=\frac{1}{3n}\). We need \(\frac{1}{3n}<\frac{1}{50}\). Both denominators are positive, so taking reciprocals reverses the comparison: \(3n>50\). Hence \(n>\frac{50}{3}=16.67\), and the smallest integer n is 17. Check the boundary: the 16th term is \(\frac1{48}\), which is greater than \(\frac1{50}\); the 17th is \(\frac1{51}\), which is smaller. Therefore B is correct. Option A fails because \(1/48>1/50\). Options C and D are later terms, so they are smaller but not the first such term. Common mistake: choose the first integer satisfying the strict inequality, not just any later index.

Given \(a_n=\frac{2n-1}{3n+1}\). Substituting \(n=8\), \(a_8=\frac{2(8)-1}{3(8)+1}=\frac{15}{25}=\frac{3}{5}\). Hence, option B is correct. A value such as \(\frac{5}{9}\) may result from incorrect substitution or simplification. Exam tip: substitute \(n\) carefully in both the numerator and denominator, then reduce the fraction.

Given \(a_n=n^2+4\), set the term equal to 260: \(n^2+4=260\). Thus, \(n^2=256=16^2\), so \(n=16\). Since a term number in a sequence is positive, \(n=-16\) is not considered. The 15th term is \(15^2+4=229\), so it is not correct. Exam tip: quickly recognising perfect squares such as 256 saves time.

The direct answer is option D, 58. Start with 2: add 3 to get 5, then multiply by 2 to get 10; add 3 to get 13, multiply by 2 to get 26; add 3 to get 29, so the next operation is multiplication by 2. Thus, 29 × 2 = 58. Option A, 32, would come from adding 3 again, but that is not the next operation. Option B, 52, does not follow either operation. Option C, 55, is also an unsupported result. Option D is correct because it follows the alternating add-3, multiply-by-2 rule. Memory cue: after every odd-position operation here, check whether the next step is multiplication.

The rule is: next term = twice the previous term + 2. Therefore, the term after 78 is \(2\times 78+2=156+2=158\). Although 156 is double of 78, the additional 2 must also be included. Exam tip: for a recurrence sequence, apply the stated rule directly to the last given term.

Equal differences give (2x+1-(x+2)=4x-5-(2x+1)), so (x=5). Exam tip: set the two consecutive differences equal.

Given \(a_n=n^2+kn\). Substituting \(n=3\), \(a_3=3^2+3k=30\), so \(9+3k=30\). Hence, \(k=7\). Now, for \(n=6\), \(a_6=6^2+6(7)=36+42=78\). Therefore, option C is correct. The value \(84\) would result from an incorrect calculation. Exam tip: first use the given term to find the unknown constant, then substitute the required value of \(n\).

In option C, the consecutive differences are \(3,5,7\), so they are not constant; the ratios are not constant either. Hence it is neither AP nor GP. Option D is constant and is both AP and GP. Exam tip: check differences first, then ratios.

The terms at odd positions are 1, 2, 4, 8, 16, \ldots, so each successive odd-position term is double the previous one. The 9th position is odd and is the fifth term of this subsequence; hence the 9th term is 16. The value 64 is the 12th term, and may be chosen by incorrectly treating the entire sequence as a single doubling pattern. Exam tip: For such sequences, list the odd-position and even-position terms separately before identifying the rule.

The successive differences are consecutive odd numbers: \(1,3,5,7,9\). Therefore, the next difference is \(11\). Hence, the next term is \(27+11=38\). The answer \(36\) would result from adding \(9\) again, but the differences continue as increasing odd numbers. Exam tip: write the consecutive differences first to identify a sequence pattern quickly.

Substitute \(n=9\): \(a_9=9^2+(-1)^9\). Since 9 is odd, \((-1)^9=-1\). Therefore, \(a_9=81-1=80\), so option C is correct. Option 82 would result from incorrectly taking \((-1)^9\) as \(+1\). Exam tip: \((-1)^n\) equals \(+1\) for even \(n\) and \(-1\) for odd \(n\).

This is an arithmetic progression with first term \(a=75\) and common difference \(d=-7\). Its \(n\)th term is \(a_n=75-7(n-1)\). For a term to be positive, \(75-7(n-1)>0\), which gives \(n<\frac{82}{7}\approx11.71\). Hence, the greatest integral value of \(n\) is 11, so 11 terms are positive. The 12th term is \(75-7\times11=-2\), so 12 terms cannot be positive. Exam tip: To count positive terms, use \(a_n>0\) and take the greatest integer value satisfying the inequality.

The first four terms are 3, 9, 27, and 81. Therefore, their sum is \(3+9+27+81=120\). This is also a geometric sequence, since each term is 3 times the previous term. Option 114 is incorrect because it is less than the actual total of the four given terms. Exam tip: when only a few terms are given, list them and add directly to avoid mistakes.

The consecutive differences are \(8-1=7\), \(19-8=11\), \(34-19=15\), and \(53-34=19\). These differences increase by \(4\) each time, so the next difference is \(23\). Therefore, the next term is \(53+23=76\). Choosing \(74\) would give a difference of \(21\), which does not follow the given difference pattern. Exam tip: For such sequences, first list the consecutive differences and check their pattern.

Given \(a_n=n^2+2^n\), \(a_4=4^2+2^4=16+16=32\) and \(a_5=5^2+2^5=25+32=57\). Therefore, \(a_4+a_5=32+57=89\). Option 85 may result from an error in calculating either \(5^2\) or \(2^5\). Exam tip: evaluate the square and exponential parts separately before adding them.

The sum of the first (6) terms is (126), and the average is (\frac{126}{6}=21). Exam tip: divide the sum by the number of terms.

Given (a_n=n!+n), substitute (n=5): (a_5=5!+5=120+5=125). Therefore, 125 is the correct option. 120 is only the value of (5!), but the additional 5 must also be added. Exam tip: in expressions involving factorials, evaluate the factorial first and then perform the remaining operations.

Given \(a_n=\frac{n^2+1}{n}\). Substituting \(n=5\), we get \(a_5=\frac{5^2+1}{5}=\frac{25+1}{5}=\frac{26}{5}\). Option B uses only \(\frac{25}{5}\) and misses the \(+1\) in the numerator. Exam tip: substitute the value of \(n\) in every part of the formula before simplifying the numerator and denominator.