nth term

Given \(a_n=6n+11\), \(a_6=6\times6+11=47\) and \(a_4=6\times4+11=35\). Therefore, \(a_6+a_4=47+35=82\). A value such as 80 may result from omitting the constant term 11 in a step. Exam tip: Substitute each value of \(n\) separately first, then add the resulting terms.

This is an arithmetic progression with first term 16 and common difference 6. Therefore, \(a_{17}=16+(17-1)\times6=16+96=112\). Hence, 112 is correct. Getting 116 would mean counting the number of common differences incorrectly. Exam tip: when finding the \(n\)th term, use \(a_n=a+(n-1)d\), not \(a+nd\).

Given \(a_n=7^n\). For the second term, substitute \(n=2\): \(a_2=7^2=49\). Therefore, 49 is correct. The value 14 comes from \(7\times2\), but here \(n\) is an exponent, not a multiplier. Exam tip: To find a particular term from \(a_n\), substitute the term number for \(n\) and identify the operation carefully.

These are odd numbers starting from (5), so (a_n=2n+3). For (n=1), the first term should be (5).

To find the fourth term, substitute \(n=4\) in the rule: \(a_4=6(4)^2-5=6\times16-5=96-5=91\). Therefore, 91 is correct. The close distractor 96 is only \(6\times16\); the subtraction of 5 is still required. Exam tip: evaluate the exponent first, then multiply and subtract.

The common ratio is \(\frac{1}{2}\), and \(a_6=144\left(\frac{1}{2}\right)^5=\frac{9}{2}\). Use powers carefully with a fractional ratio.

To find the ninth term, substitute 9 for n in the formula: \(a_9=13\times9-6=117-6=111\). Therefore, the correct answer is 111. The value 117 is only \(13\times9\); subtracting 6 is also necessary. Exam tip: In any nth-term question, substitute the given value of n carefully before simplifying.

The first term is (28) and the difference is (-6), so (a_n=28+(n-1)(-6)=34-6n). Pay attention to the sign in a decreasing sequence.

Given \(a_n=n(n+3)\). Substituting \(n=6\), we get \(a_6=6(6+3)=6\times9=54\). Therefore, 54 is correct. The value 48 would result from incorrectly using 8 in place of \(n+3\). In exams, substitute the value of \(n\) into the entire expression before simplifying.

The common ratio is (3), so (a_4=8\cdot3^3=216). In a geometric term, the exponent of the ratio is (n-1).

Given \(a_n=9n+5\), \(a_9=9\times9+5=86\) and \(a_6=9\times6+5=59\). Hence, \(a_9-a_6=86-59=27\). Option 24 is incorrect because the difference between the indices is \(9-6=3\), and each successive term increases by 9; thus the difference is \(3\times9=27\). Exam tip: for a linear sequence \(a_n=dn+c\), use \(a_p-a_q=d(p-q)\) for a quick calculation.

The sequence is 9, 15, 23, 33. Substituting \(n=1,2,3,4\) in \(n^2+3n+5\) gives 9, 15, 23, and 33 respectively, so it is the required \(n\)th term. Although \(n^2+8\) looks similar, it gives 12 as the second term. Exam tip: verify a proposed nth-term rule by checking at least the first three terms.

To find the zero term, put \(a_n=0\). Thus, \(72-8n=0\Rightarrow 8n=72\Rightarrow n=9\). Therefore, the 9th term is zero. The 8th term is not zero because \(a_8=72-8(8)=8\). Exam tip: To find the position of a required term, substitute its given value in \(a_n\) and solve for \(n\).

The terms are (4,16,64,256,1024), so (1024) is the fifth term. For small questions, listing terms is quick.

Given \(a_n=\frac{n(n+5)}{2}\), substitute \(n=7\): \(a_7=\frac{7(7+5)}{2}=\frac{7\times12}{2}=7\times6=42\). Hence, 42 is the correct option. The value 48 could result from incorrectly taking \(7+5\) as 14. Exam tip: When finding a term of a sequence, substitute the given value of \(n\) carefully and simplify the bracket first.

This is an arithmetic progression because each term decreases by 4. Here, \(a=35\), \(d=-4\), and \(n=10\). Therefore, \(a_{10}=a+(n-1)d=35+9(-4)=35-36=-1\). Hence, -1 is correct. Getting 1 is a common error caused by applying the common difference incorrectly instead of using the 9 gaps before the 10th term. Exam tip: for the \(n\)th term, always use \(n-1\) common differences.

Given (a_n=5n^2-4n), substitute n=5: (a_5=5(5)^2-4(5)=5×25-20=125-20=105). Hence, 105 is the correct answer. An answer such as 100 may result from an error while squaring or multiplying. Exam tip: after substitution, evaluate powers first, then multiplication, and finally subtraction.

The successive differences are \(6,8,10\), whose differences are constant at \(2\); therefore, the term should contain an \(n^2\) part. For \(a_n=n^2+3n\), we get \(a_1=4\), \(a_2=10\), \(a_3=18\), and \(a_4=28\). The close distractor \(n^2+4\) gives 5 as its first term, so it is incorrect. Exam tip: test a proposed nth-term formula by substituting \(n=1\) and \(n=2\).

Given \(a_n=4n+18\) and \(a_n=70\), we get \(4n+18=70\). Thus, \(4n=52\), so \(n=13\). Therefore, 70 is the 13th term of the sequence. If \(n=12\), the term is \(4(12)+18=66\), not 70. Exam tip: To find a term’s position, substitute its value for \(a_n\) and solve the equation for \(n\).

The common ratio is \(\frac{1}{3}\), and \(a_5=120\left(\frac{1}{3}\right)^4=\frac{40}{27}\). Use powers carefully with a fractional ratio.