Sequences

The general term is (\frac{n}{n+1}), so the (10)th term is (\frac{10}{11}). Exam tip: observe numerator and denominator patterns separately.

The general term is (\frac{n}{2n+1}), so the (8)th term is (\frac{8}{17}). Exam tip: identify the denominator pattern separately.

(\frac{1}{21}<\frac{1}{20}), while (\frac{1}{20}) is not smaller. Exam tip: strict inequality does not include equality.

We need \(\frac{n}{n+2}>\frac45\). Since \(5(n+2)\) is positive, cross-multiplication gives \(5n>4n+8\), so \(n>8\). The smallest positive integer greater than 8 is 9; hence the first such term is \(a_9\). At \(n=8\), \(a_8=\frac{8}{10}=\frac45\), which is equal to, not greater than, \(\frac45\). Exam tip: for “greater than”, do not include the equality case.

After the fifth term, (n=5), so (2\times57+6=120). Exam tip: use the correct current value of (n) in recursion.

The recurrence requires adding consecutive squares: \(a_2=3+1^2=4\), \(a_3=4+2^2=8\), \(a_4=8+3^2=17\), \(a_5=17+4^2=33\), and \(a_6=33+5^2=58\). Hence, 58 is correct. Option 56 is incorrect because the correct sum \(1^2+2^2+3^2+4^2+5^2=55\) must be added to 3. Exam tip: to find \(a_6\), add squares from \(1^2\) through \(5^2\).

Given a_n=2n^3+1, a_5=2(5^3)+1=2(125)+1=251 and a_6=2(6^3)+1=2(216)+1=433. Therefore, a_5+a_6=251+433=684. An answer such as 680 can result from an error while evaluating a cube or adding the terms. Exam tip: calculate each required term separately before adding them.

Given \(a_n=5n-(-1)^n\). Substituting \(n=8\), \(a_8=5\times8-(-1)^8=40-1=39\), since an even power of \((-1)\) equals \(1\). Option 41 would result from incorrectly taking \((-1)^8\) as \(-1\). Exam tip: remember that \((-1)^{\text{even}}=1\) and \((-1)^{\text{odd}}=-1\).

The terms are \(1^3+1=2\), \(2^3+1=9\), \(3^3+1=28\), and \(4^3+1=65\). Therefore, the next term is \(5^3+1=125+1=126\). Although 128 is close, it equals \(5^3+3\), not the stated rule. Exam tip: verify a sequence rule by substituting \(n=1,2,3\) into it.

Constant non-zero second differences indicate a quadratic sequence, typically with a term of the form \(an^2+bn+c\). An arithmetic progression has constant first differences instead. Exam tip: use a difference table to classify sequences.

In A, the first differences are \(5, 7, 9\), giving equal non-zero second differences \(2, 2\). In B, the first difference itself is constant, so it is an arithmetic sequence. Exam tip: check differences twice.

The direct answer is option A, \(x=5\). Equal consecutive differences mean the difference from the first term to the second equals the difference from the second to the third. Thus \((x+3)-x= (2x+1)-(x+3)\). The left side is 3, and the right side is \(x-2\). So \(3=x-2\), giving \(x=5\). Check: the terms become 5, 8, 11, whose differences are 3 and 3. Option A works. Option B, x=6, gives 6, 9, 13, with differences 3 and 4. Option C, x=7, gives 7, 10, 15, with differences 3 and 5. Option D, x=8, gives 8, 11, 17, with differences 3 and 6. Therefore only A satisfies the condition. For three consecutive arithmetic terms, always equate the two neighboring differences.

Given \(a_n=kn+2\). Substituting \(n=5\), \(a_5=5k+2=27\), so \(5k=25\) and \(k=5\). Now, for \(n=9\), \(a_9=9\times5+2=47\). Hence, 47 is correct. The value 45 would result from incorrectly omitting the constant term \(+2\). Exam tip: first find the unknown constant from the given term, then substitute the required value of \(n\).

Given \(a_n=pn^2+q\), we have \(a_1=p+q=5\) and \(a_2=4p+q=14\). Subtracting the first equation from the second gives \(3p=9\), so \(p=3\) and \(q=2\). Therefore, \(a_4=3\times4^2+2=3\times16+2=50\). Hence, 50 is the correct answer. Although 48 is a close distractor, it does not result from correctly using \(4^2=16\). Exam tip: first find the unknown constants from the given terms, then substitute the required value of \(n\).

Given \(a_n=n^2+cn\), substitute \(n=4\): \(a_4=4^2+4c=32\). Thus, \(16+4c=32\), so \(c=4\). Now substitute \(n=9\): \(a_9=9^2+4\times9=81+36=117\). Therefore, the correct answer is 117. Option 114 is close, but it is 3 less than the correct value of \(a_9\). Exam tip: first use the given term to find the unknown constant, then substitute the required value of \(n\).

Here (a_n=n(n+1)), and (9\times10=90) while (10\times11=110). So (9) terms are less than (100).

Given \(a_n=2^n+n^2\), substitute \(n=5\): \(a_5=2^5+5^2=32+25=57\). Therefore, 57 is the correct option. An answer such as 55 can result from an error in evaluating the power or the square. Exam tip: substitute the value of \(n\) into every part of the formula, then calculate powers and squares separately.

Since (\frac{10\times13}{2}=65), it is the (10)th term. Exam tip: substitute options directly into the formula.

The differences between consecutive terms are 3, 4, 7, and 12. Their second differences are 1, 3, and 5, so the next second difference is 7. Hence, the next first difference is 12+7=19, and the next term is 28+19=47. If 45 were chosen, the difference would be only 17, not the required 19. Exam tip: For such sequences, list first differences and then second differences to identify the pattern.

For \(a_n=n^2+1\), the first difference is \(2n+1\), so it changes with \(n\), while the second difference is the constant \(2\). A linear sequence has constant first differences instead. Exam tip: check second differences when an \(n^2\) term appears.

Here the useful pattern is not in the terms themselves but in the differences between consecutive terms. From 2 to 7 the increase is 5, from 7 to 17 it is 10, from 17 to 32 it is 15, and from 32 to 52 it is 20. These differences increase by 5 each time. The next difference must therefore be 25. Adding that difference to the last known term gives 52 plus 25 equals 77, so option C is correct.

The calculation should be performed in two stages: first extend the difference pattern, then apply it to the last term. The difference list is 5, 10, 15, 20, 25, so the next term is 52 + 25. This equals 77. A value of 72 would add only 20 again, while 75 or 80 does not follow the next difference in the stated pattern. The supplied answer C is therefore consistent, and its explanation correctly emphasizes that recognizing the successive differences is the key step.

Odd positions are (4,8,16,\ldots), so the next odd-position term is (32). Exam tip: separate odd and even positions in alternating sequences.

Apply the recurrence step by step: \(a_2=3(2)-1=5\), \(a_3=3(5)-1=14\), and \(a_4=3(14)-1=41\). Therefore, \(a_5=3(41)-1=122\). The value 116 can result from an error in multiplication or subtraction in the final step. Exam tip: write each intermediate term before finding the required term.

The direct answer is option D, \(a_5=41\). Start with \(a_1=1\) and use \(a_{n+1}=a_n+n(n+1)\) carefully. For n=1, \(a_2=a_1+1\cdot2=1+2=3\). For n=2, \(a_3=3+2\cdot3=9\). For n=3, \(a_4=9+3\cdot4=21\). For n=4, \(a_5=21+4\cdot5=41\). Thus the terms are 1, 3, 9, 21, 41. Option A, 31, results from an arithmetic-style shortcut and is wrong. Option B, 36, misses the final addition or uses an incorrect product. Option C, 39, also does not follow the recurrence. Option D matches the complete step-by-step calculation. The important exam point is that to obtain \(a_{n+1}\), use the current index n, not n+1, in \(n(n+1)\).

The general term is (\frac{n+1}{2n+1}), so the (12)th term is (\frac{13}{25}). Exam tip: observe numerator and denominator patterns separately.