nth term

Given \(a_n=7n-1\), set the term equal to 69: \(7n-1=69\). Thus, \(7n=70\), so \(n=10\). Hence, 69 is the 10th term of the sequence. The 9th term is \(7\times9-1=62\), so it is not correct. Exam tip: To find the position of a given term, equate \(a_n\) to that value and solve for \(n\).

This is an arithmetic progression with first term 5 and common difference 7. Therefore, \(a_{13}=a+(13-1)d=5+12\times7=89\). Hence, the correct answer is 89. A value such as 87 can result from incorrectly counting the number of common differences. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

The direct answer is A: \(7\cdot3^{n-1}\). In a geometric progression, the same multiplier is used each time. The first term is \(a=7\), and the common ratio is \(r=21/7=3\); the next check gives \(63/21=3\). Therefore the nth term is \(a_n=ar^{n-1}=7\cdot3^{n-1}\). The exponent is \(n-1\), not \(n\), because the first term has not yet been multiplied by the ratio. Checking terms confirms the rule: at \(n=1\), \(7\cdot3^0=7\); at \(n=2\), \(7\cdot3=21\); at \(n=3\), \(7\cdot3^2=63\). Option A matches all of these facts. Option B reverses the first term and ratio, so it starts with 3 rather than 7. Option C is linear and gives values such as 10 when \(n=1\), not 7. Option D starts with 21, so it gives the second term at \(n=1\), not the first. Memory cue: first term stays outside; common ratio is raised to one less than the term number.

To find the seventh term, substitute \(n=7\) in the formula: \(a_7=7^2+3\times7=49+21=70\). Therefore, the correct answer is 70. Getting 77 may result from an error in multiplication or addition. Exam tip: after substituting the term number, evaluate powers and multiplication before addition.

The first term is (21) and the difference is (-5), so (a_n=21+(n-1)(-5)=26-5n). In a decreasing sequence, the difference is negative.

Given \(a_n=55-4n\), substitute \(n=11\): \(a_{11}=55-4(11)=55-44=11\). Hence, 11 is the correct answer. Getting 9 would require an incorrect multiplication or subtraction. Exam tip: after substituting the value of \(n\), multiply first and then subtract.

Given \(a_n=10n+1\), \(a_8=10\times8+1=81\) and \(a_3=10\times3+1=31\). Therefore, \(a_8-a_3=81-31=50\). Option 45 may result from using only the index difference \(8-3\), but each increase of 1 in the index increases the term by 10. Exam tip: find both terms, or directly use \(a_8-a_3=10(8-3)\).

This is an arithmetic progression with first term 10 and common difference 7. Its nth term is \(a_n=10+(n-1)\times 7=7n+3\). Setting \(7n+3=73\) gives \(7n=70\), so \(n=10\). Therefore, 73 is the 10th term. The 9th term is 66, so it is a close but incorrect option. Exam tip: To find the position of a given term, form \(a_n\) first and equate it to that term.

Given \(a_n=6^n\), substitute \(n=3\) to find the third term: \(a_3=6^3=6\times6\times6=216\). Hence, 216 is correct. The value 36 is \(6^2\), so it is not the third term. Exam tip: To find an \(n\)th term, substitute the required value of \(n\) carefully into the formula.

The correct term is \(a_n=n^2+2n+5\). Substituting \(n=1,2,3,4\) gives \(8,13,20,29\), respectively. In option B, the second term is \(11\), while in option D it is \(14\), so neither matches the sequence. Exam tip: For sequences with changing first differences, verify a possible quadratic expression using the first few values of \(n\).

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 direct answer is option C, 216. This is a geometric progression because the common ratio is \(24\div8=3\), and \(72\div24=3\). The nth-term formula is \(a_n=ar^{n-1}\). For the fourth term, \(a_4=8 imes3^{4-1}=8 imes3^3=8 imes27=216\). Direct continuation gives the same result: 8, 24, 72, then \(72 imes3=216\). Option A, 144, is not three times 72 and does not follow the pattern. Option B, 196, also has no correct relation to the common ratio. Option C, 216, is correct because it is the next term after multiplying 72 by 3. Option D, 288, is too large and results from an incorrect multiplication or exponent. Remember: count the jumps from the first term; to reach the fourth term, multiply by the ratio three times.

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.