nth term

Putting \(n=1,2,3\) gives 5, 9, 13. The consecutive differences are \(9-5=4\) and \(13-9=4\), so it is an AP with common difference 4, not a GP. Exam tip: compare consecutive differences to identify an AP.

This is an arithmetic progression with first term 12 and common difference 12. Therefore, \(a_7=12+(7-1)\times 12=84\). The number 72 is the sixth term, so it is a close but incorrect option. In exams, use \(a_n=a+(n-1)d\) instead of counting terms one by one.

For the ninth term, substitute n=9: a_9=18-9=9. Therefore, the correct answer is 9. The value 8 would result from using n=10, so the term number must be substituted carefully. Exam tip: to find the nth term, directly replace n in the given formula with the required term number.

The governing concept is the general term of an arithmetic progression. The first term is a₁ = 7, and the common difference is d = 11 − 7 = 4; the next differences are also 4. Apply aₙ = a₁ + (n − 1)d: aₙ = 7 + (n − 1)4 = 7 + 4n − 4 = 4n + 3. Thus option A is correct. Substituting n = 1 gives 7, n = 2 gives 11, n = 3 gives 15, and n = 4 gives 19, confirming the rule. Option B uses an unsuitable coefficient and begins at 4, option C gives 1 for the first term, and option D has common difference 1 rather than 4. Therefore, the other choices are eliminated by direct checking.

For an AP, \(a_n=a+(n-1)d\). With \(a=3\) and \(d=4\), \(a_n=3+4(n-1)=4n-1\). Option C gives first term 7, not 3. Exam tip: put \(n=1\) to check the first term.

This is the sequence of odd numbers, whose nth term is \(a_n=2n-1\). Putting \(2n-1=31\) gives \(2n=32\), so \(n=16\). Therefore, 31 is the sixteenth term. The fifteenth term is \(2(15)-1=29\), so it is not correct. Exam tip: for a sequence of odd numbers, use \(2n-1\) as the nth-term formula.

For the third term, substitute \(n=3\): \(a_3=3^3+1=27+1=28\). Therefore, the correct answer is \(28\). The nearby option \(27\) is only the value of \(3^3\); the \(+1\) still has to be added. Exam tip: while finding a term, substitute the value of \(n\) first and then complete the operations in order.

Dividing the given terms by 3 gives \(1,4,9,16\), which are \(1^2,2^2,3^2,4^2\) respectively. Therefore, the \(n\)th term is \(a_n=3n^2\). For \(a_n=3n\), the second term would be 6, not 12. Exam tip: Check early terms for a common multiplier and square or cube patterns.

Given \(a_n=\frac{n+3}{2}\). To find \(a_7\), substitute \(n=7\): \(a_7=\frac{7+3}{2}=\frac{10}{2}=5\). Therefore, 5 is the correct option. The value 4 would result only if the numerator were 8, so it is not correct here. Exam tip: for an nth-term formula, substitute the given value of \(n\) before simplifying.

At (n=1), it gives (2), and at (n=2), it gives (\frac{5}{2}). In exams, match a fractional rule with the first two terms.

To find the second term, substitute \(n=2\): \(a_2=4^2=4\times4=16\). Therefore, the correct answer is 16. The value 8 comes from multiplying 4 by 2, not from evaluating \(4^2\). Exam tip: first substitute the required value of n in the formula, then calculate the exponent carefully.

The formula for an AP is \(a_n=a+(n-1)d\). Substituting \(a=4\) and \(d=3\) gives \(a_n=4+3(n-1)\). In option B, putting \(n=1\) gives 7, not the first term 4. Exam tip: test a formula with \(n=1\).

The governing concept is the explicit nth-term rule for an arithmetic progression. The first term is a₁ = 6, and the common difference is d = 13 − 6 = 7; both later differences are also 7. Using aₙ = a₁ + (n − 1)d gives aₙ = 6 + 7(n − 1) = 6 + 7n − 7 = 7n − 1. Hence option A is correct. Checking the first indices gives a₁ = 7 − 1 = 6, a₂ = 14 − 1 = 13, and a₃ = 21 − 1 = 20; n = 4 gives 27 as well. Option B starts at 8, option C uses 6 instead of the required difference 7 as the coefficient of n, and option D increases by only 1.

\(a_{n+1}-a_n=[4(n+1)+1]-(4n+1)=4\) is constant for every \(n\), so the sequence is an arithmetic progression. A GP needs a constant ratio instead. Exam tip: check consecutive differences first.

This is an arithmetic progression with first term 20 and common difference 20. Therefore, its nth term is \(a_n=20n\). From \(20n=160\), we get \(n=8\), so 160 is the eighth term. The seventh term is 140, so it is not correct. Exam tip: To find a term number, use the nth-term formula or identify the multiple of the common factor.

For the fourth term, substitute \(n=4\): \(a_4=\frac{5\times4}{2}=\frac{20}{2}=10\). Therefore, the correct answer is 10. Option 12 would result from an incorrect division of 20. Exam tip: in an nth-term formula, substitute the value of \(n\) first and then simplify step by step.

This is an arithmetic progression with first term \(a=\frac{5}{2}\) and common difference \(d=\frac{5}{2}\). Therefore, \(a_n=a+(n-1)d=\frac{5}{2}+(n-1)\frac{5}{2}=\frac{5n}{2}\). In option B, putting \(n=1\) gives 5, whereas the first term is \(\frac{5}{2}\). Exam tip: after choosing a formula, substitute \(n=1\) to check whether it gives the first term.

For the third term, put n=3: a_3=3^3+2=27+2=29. Therefore, 29 is correct. 27 is only the value of 3^3 and misses the given +2. Exam tip: To find an nth term, substitute the value of n first, then follow the order of operations.

For an AP, \(a_n=a+(n-1)d\). Substituting \(a=7\) and \(d=3\) gives \(a_n=7+3(n-1)\). In option B, putting \(n=1\) gives 10, not the first term 7. Exam tip: always test a formula using \(n=1\).

The governing concept is locating a term by using the general rule of the sequence. Every displayed term is a consecutive multiple of 15: 15 = 15×1, 30 = 15×2, 45 = 15×3, and 60 = 15×4. Therefore, the nth term is aₙ = 15n. To find the position of 105, solve 15n = 105. Dividing both sides by 15 gives n = 7, so 105 is the seventh term and option C is correct. A neighboring-term check confirms the result: the fifth term is 75, the sixth is 90, the seventh is 105, and the eighth is 120. Thus options A, B, and D identify values different from 105.

For the third term, substitute \(n=3\): \(a_3=3(3+2)=3\times5=15\). Therefore, the correct answer is \(15\). A value such as \(18\) may result from an error in evaluating \(n+2\) or in multiplication. Exam tip: substitute the term number correctly for \(n\), then simplify step by step.

For the sequence \(3,8,15,24\), substituting \(n=1,2,3,4\) in \(a_n=n(n+2)\) gives \(3,8,15,24\), respectively. Hence, \(a_n=n(n+2)\) is correct. The close distractor \(n(n+1)+1\) gives the first term as 3, but its second term is 7, not 8. Exam tip: verify an nth-term formula using at least the first two or three terms.

For an AP, \(a_n=a+(n-1)d\). Here \(a=4\) and \(d=3\), so \(a_n=4+3(n-1)\). Putting \(n=1\) gives 4; this is a quick exam check for the first term.

Each term in the sequence is a successive multiple of 11, so its nth term is \(a_n=11n\). Putting \(11n=99\) gives \(n=9\). Therefore, 99 is the ninth term. The eighth term is 88, so it is not correct. Exam tip: for such sequences, divide the given term by the common multiple to check its position.

Since \(a_{n+1}-a_n=5(n+1)-2-(5n-2)=5\), the difference between consecutive terms is constant. Hence it is an AP with common difference 5. In exams, check \(a_{n+1}-a_n\) to identify an AP.