Sequences

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 governing concept is using successive differences to extend a sequence. Subtracting consecutive terms gives 6 − 4 = 2, 9 − 6 = 3, 14 − 9 = 5, 21 − 14 = 7, and 32 − 21 = 11. These are consecutive prime numbers. The next prime after 11 is 13, so the next term must be the current final term plus 13: 32 + 13 = 45. Therefore option C is correct. The distractor 41 would add 9, 43 would add 11 by repeating the previous difference, and 47 would add 15; none follows the stated prime-difference pattern. Writing the differences separately avoids confusing terms with increments.

After (29), multiplication by (2) is applied, so (29\times2=58). Exam tip: identify the next operation in alternating-operation sequences.

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.

Use the recurrence step by step. Thus, a_2=2(2)+1^2=5, a_3=2(5)+2^2=14, and a_4=2(14)+3^2=37. Therefore, 37 is correct. Although 39 is a close distractor, it is not the correct value of 2(14)+3^2. Exam tip: to find a_4, substitute n=1, then n=2, and then n=3; do not substitute n=4 directly.

In option B, the sequence falls from 1 to -1 and then rises from -1 to 1, so it does not move in one direction. A and D increase, while C decreases. Exam tip: compare consecutive terms.