Sequences

(a_{12}=192) and (a_{13}=221), so (12) terms are less than (200). Exam tip: check the two terms near the boundary.

The question asks for the sum of the first five terms, so the listed terms from the first through the fifth must all be included exactly once. They are 2, 5, 10, 17, and 26. This is not asking for the fifth term alone. One may notice that the terms follow the rule \(n^2+1\), but direct addition is the simplest method for the requested small number of terms.

Add the terms in order: \(2+5=7\), \(7+10=17\), \(17+17=34\), and \(34+26=60\). Therefore the sum is 60, so option D is correct. Omitting the first term would give an incorrect total, and stopping at the fourth term would give only 34. Every one of the five specified terms must be counted.

Successive differences mean that each new term is obtained by adding the stated difference to the preceding term. Thus, \(a_2=2+3\), \(a_3=a_2+7\), \(a_4=a_3+11\), and \(a_5=a_4+15\). Therefore, \(a_5=2+3+7+11+15=38\). Option 40 would result from incorrectly adding 2 extra to the total. Exam tip: to find \(a_n\), add the first \(n-1\) successive differences to \(a_1\).

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.

Equal differences give (x+3-x=2x+1-(x+3)), so (x=5). Exam tip: set the two consecutive differences equal.

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.