nth term

This is an arithmetic progression because \(a_{n+1}-a_n=[6(n+1)+1]-(6n+1)=6\), a constant difference. A geometric progression requires a constant ratio instead. Exam tip: check consecutive-term differences first.

The direct answer is option A, \(a_n=6n\). Look at the terms: 6, 12, 18, and 24. They are respectively \(6\times1\), \(6\times2\), \(6\times3\), and \(6\times4\). Therefore the nth term is \(6\times n=6n\). This is also an arithmetic progression with first term 6 and common difference 6: \(a_n=6+(n-1)6=6n\). Option A is correct because it gives 6 for \(n=1\), 12 for \(n=2\), and so on. Option B, \(n+6\), gives 7 when \(n=1\), so it does not even give the first term. Option C, \(12n\), gives 12 for the first term, which is too large. Option D, \(6n-1\), gives 5 for the first term and therefore fails. The memory trick is: when the sequence is 6 times 1, 2, 3, 4, its nth term is 6 times n.

For the sixth term, substitute \(n=6\). Thus, \(a_6=6^2-2=36-2=34\). Therefore, 34 is correct. The close distractor 36 is only \(6^2\); the subtraction of 2 has not been performed. Exam tip: after substituting the value of \(n\), evaluate the power first and then subtract.

Direct answer: aₙ = 3n − 2, so B is correct. The differences are constant: 4 − 1 = 3, 7 − 4 = 3, and 10 − 7 = 3. Therefore the sequence is arithmetic, with first term a₁ = 1 and common difference d = 3. Use aₙ = a₁ + (n − 1)d. Substitution gives aₙ = 1 + (n − 1)3 = 1 + 3n − 3 = 3n − 2. Verify it at several positions: n = 1 gives 1, n = 2 gives 4, n = 3 gives 7, and n = 4 gives 10. A starts at 4, so it is incorrect. C has difference 1, not 3. D starts at 1 but has difference 4, so it matches neither the second term nor the pattern. Common confusion: the coefficient of n is the common difference, but the constant term must be adjusted so that n = 1 gives the actual first term.

In \(a_n=5n+2\), the coefficient of \(n\) is 5, so each successive term increases by 5. Hence it is an arithmetic sequence with common difference 5. A geometric sequence requires a constant ratio between consecutive terms. Exam tip: in \(a_n=dn+c\), identify \(d\) as the common difference.

This sequence consists of multiples of 7, so its nth term is \(a_n=7n\). Putting \(7n=70\) gives \(n=10\). Therefore, 70 is the tenth term. The ninth term is 63, so option B is not correct. Exam tip: for this sequence, divide the given number by 7 to find its term number.

For the seventh term, substitute 7 for n: \(a_7=12-7=5\). Therefore, the correct answer is 5. Getting 6 would result from subtracting the wrong term number from 12. Exam tip: to find the nth term, replace n with the required term number.

This is an arithmetic sequence because 3 is added to each successive term. Starting with 2, the sixth term is obtained by adding 3 five times: \(2+5\times3=17\). Therefore, 17 is correct. The value 16 would result from adding 3 only four times. Exam tip: to find the \(n\)th term, add the common difference \(n-1\) times to the first term.

This is an arithmetic sequence because each successive term increases by 3. For the first term, \(a_1=3(1)+1=4\), so \(a_n=3n+1\) is correct. The formula \(3n-1\) gives 2 as the first term. Exam tip: verify a formula using the first two terms.

For the fourth term, substitute \(n=4\). Thus, \(a_4=4^3=4\times4\times4=64\). Therefore, \(64\) is the correct answer. \(16\) is a close distractor because it equals \(4^2\), not \(4^3\). Exam tip: \(n^3\) means multiplying \(n\) by itself three times.

This is an arithmetic progression with first term \(a=8\) and common difference \(d=8\). Thus, \(a_{10}=a+(10-1)d=8+9\times8=80\). The value 72 is the ninth term, so it is a close but incorrect option. In exams, use \(a_n=a+(n-1)d\) for the nth term of an AP.

\(a_n=4n+1\) has the linear form \(pn+q\), so consecutive terms differ by the constant \(4\); hence it is an arithmetic progression. For \(n^2+1\), the differences change. Exam tip: check whether \(a_{n+1}-a_n\) is constant.

Direct answer: aₙ = 5n − 2, so A is correct. Subtract consecutive terms: 8 − 3 = 5, 13 − 8 = 5, and 18 − 13 = 5. The constant difference is 5, so this is an arithmetic progression. Its first term is a₁ = 3. Using aₙ = a₁ + (n − 1)d, we get aₙ = 3 + (n − 1)5 = 3 + 5n − 5 = 5n − 2. Verify: n = 1 gives 5 − 2 = 3; n = 2 gives 10 − 2 = 8; n = 3 gives 15 − 2 = 13; and n = 4 gives 20 − 2 = 18. B gives 7 at the first position, so it fails immediately. C gives 8 at n = 1 and changes by 3, while D gives 6 at n = 1 and changes by 1. Memory cue: in an arithmetic sequence, start with the first term and add the common difference n − 1 times, not n times.

Putting \(n=1\) gives the first term \(3(1)-2=1\), and \(n=2\) gives \(3(2)-2=4\). Hence the sequence is 1, 4, 7, 10, .... In exams, verify the first two terms to identify the sequence quickly.

This sequence consists of multiples of 9, so its nth term is \(a_n=9n\). Putting \(9n=81\) gives \(n=9\). Therefore, 81 is the ninth term. The eighth term is \(9\times8=72\), so it is not correct. Exam tip: To find a term number, equate the given value to the nth-term formula and solve for \(n\).

For the sixth term, substitute \(n=6\) in the rule: \(a_6=\frac{6+2}{2}=\frac{8}{2}=4\). Therefore, the correct answer is 4. Option 3 can result from an incorrect division after adding 6 and 2. Exam tip: in \(a_n\), first substitute the given subscript for \(n\), then simplify step by step.

At (n=1) it gives (1), and at (n=2) it gives (\frac{3}{2}), so the rule is correct. In exams, match fractional sequences with terms.

For the third term, substitute n=3. Thus, a_3=3^3=27, so 27 is correct. The value 9 equals 3^2, so it is the second term, not the third. Exam tip: To find the nth term, replace n in the formula with the required term number.

This is an arithmetic progression with first term \(a=4\) and common difference \(d=9-4=5\). Its \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_7=4+(7-1)\times5=34\). Hence, 34 is correct. The value 29 is the fifth term, not the seventh term. Exam tip: identify the first term and common difference before substituting \(n\) in \(a_n=a+(n-1)d\).

Twice the position number \(n\) is \(2n\). Adding 5 gives \(a_n=2n+5\). Option \(2n-5\) incorrectly subtracts 5. Exam tip: translate “more than” as addition before choosing the formula.

This is an arithmetic progression with first term \(a=12\) and common difference \(d=12\). Thus, \(a_n=a+(n-1)d=12+(n-1)\times12=12n\). The expression \(a_n=12n+12\) gives 24 as the first term, so it is incorrect. Exam tip: Substitute \(n=1\) to check whether the rule gives the first term.

For the fourth term, substitute n=4. Thus, a_4=4^2+4=16+4=20. Therefore, 20 is the correct answer. The value 16 is only 4^2; the additional +4 must also be included. Exam tip: when finding an nth term, substitute the term number carefully in the formula.

The rule for this sequence is \(a_n=n(n+1)=n^2+n\). Substituting \(n=1,2,3,4\) gives \(2,6,12,20\), respectively. With \(a_n=n^2\), the second term would be \(4\), not the given \(6\). Exam tip: verify a proposed nth-term rule using the first few terms.

\(n\) indicates the position of a term, so \(a_4\) is the fourth term of the sequence. The value of \(a_n\) comes from its rule; the common difference is a separate quantity. Exam tip: read the subscript as the term number.

Each term in this sequence is a multiple of 10, so its nth term is \(a_n=10n\). Putting \(10n=90\) gives \(n=9\). Therefore, 90 is the ninth term. The tenth term would be 100, so it is not correct. Exam tip: for this sequence, divide the given number by 10 to find its term number quickly.