nth term

The governing concept is finding the position of a given value by using the general term. This is an arithmetic sequence with first term 3 and common difference 3, so its nth term is aₙ=3n. To find the position of 36, set 3n=36 and divide both sides by 3: n=12. Hence 36 is the twelfth term, so option C is correct. Direct listing also confirms it: the kth multiple of 3 is 3k, and 36=3×12. Option A gives 3×10=30, option B gives 3×11=33, and option D gives 3×13=39. Those values lie near 36 but do not equal it, so they are plausible counting distractors.

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.

Each term is a multiple of (6), so (a_n=6n). In exams, the rule of a multiples sequence is easy.

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.

This is an arithmetic progression because the difference between consecutive terms is constant: 4 − 1 = 3, 7 − 4 = 3, and 10 − 7 = 3. For an arithmetic progression, the nth-term formula is aₙ = a₁ + (n − 1)d, where a₁ is the first term and d is the common difference. Substituting a₁ = 1 and d = 3 gives aₙ = 1 + (n − 1)3 = 1 + 3n − 3 = 3n − 2. Therefore, option B is correct. A quick check gives a₁ = 3(1) − 2 = 1, a₂ = 4, and a₃ = 7. Option A begins with 4, while options C and D do not reproduce the given sequence.

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.

The sequence is an arithmetic progression since every term increases by 5: 8 − 3 = 5, 13 − 8 = 5, and 18 − 13 = 5. The general term of an arithmetic progression is aₙ = a₁ + (n − 1)d. Here the first term is a₁ = 3 and the common difference is d = 5. Thus aₙ = 3 + (n − 1)5 = 3 + 5n − 5 = 5n − 2. Hence option A is correct. Substitution confirms the result: n = 1 gives 3, n = 2 gives 8, n = 3 gives 13, and n = 4 gives 18. Option B gives 7 as its first term; option C gives 8, and option D gives 6, so none of those can represent the given sequence.

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 \(a_n=a+(n-1)d\).