nth term

The governing concept is the general term of a sequence. Every listed number is a multiple of 15: 15 = 15 × 1, 30 = 15 × 2, 45 = 15 × 3, and 60 = 15 × 4. Therefore, the nth term is aₙ = 15n. To locate 105, solve 15n = 105. Dividing both sides by 15 gives n = 105 ÷ 15 = 7. Hence, 105 occupies the seventh position, so option C is correct. The fifth, sixth, and eighth terms would be 75, 90, and 120 respectively, so those distractors do not equal 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.

The governing concept is the explicit nth-term formula for an arithmetic sequence. The first term is a₁ = 10, and the common difference is d = 13 − 10 = 3. For an arithmetic sequence, aₙ = a₁ + (n − 1)d. Substituting the values gives aₙ = 10 + (n − 1)3 = 10 + 3n − 3 = 3n + 7. Thus option A is correct. A quick check confirms it: at n = 1, the formula gives 3(1) + 7 = 10; at n = 2 it gives 13; and at n = 4 it gives 19. Option B has the wrong coefficient arrangement, option C gives 4 at n = 1, and option D does not have the required difference of 3.

For the fourth term, substitute n = 4: \(a_4=4^2+4+1=16+4+1=21\). Therefore, 21 is the correct answer. The value 20 would result if the constant term 1 were omitted after adding \(4^2\) and 4. Exam tip: To find an nth term, substitute the required value of n and simplify step by step.

For the given sequence, substitute \(n=1,2,3,4\) into \(a_n=n^2+n+1\). This gives \(3,7,13,21\), so option A is correct. The close distractor \(a_n=3n+1\) gives 4 as its first term, not 3. Exam tip: test a proposed nth-term formula with at least the first two values of \(n\).

In \(a_n=5-3n\), the coefficient of \(n\) is \(-3\), so every next term decreases by 3 and the common difference is \(-3\). Option B has common difference \(+3\). Exam tip: for \(a_n=pn+q\), the common difference is \(p\).

This is an arithmetic sequence with first term \(a=14\) and common difference \(d=14\). Thus, \(a_n=a+(n-1)d=14+(n-1)14=14n\). Hence, \(a_n=14n\) is correct. If \(a_n=14n+14\), the first term would be 28, so it is incorrect. Exam tip: put \(n=1\) in the formula to check whether it gives the first term.

The governing concept is direct substitution into an explicit formula. The sequence is defined by aₙ = n(n + 1)/2, and the question asks for the sixth term, so substitute n = 6. This gives a₆ = 6(6 + 1)/2 = 6 × 7/2 = 42/2 = 21. Therefore, option C is correct. The value 15 is the fifth triangular number, since 5 × 6/2 = 15. The values 18 and 24 can result from incorrect substitution or arithmetic, but neither follows from the stated formula at n = 6. The important step is to use the requested index six, not to add six to a previous term or confuse the formula with n².

Its rule is (a_n=\frac{n(n+1)}{2}), and (a_6=21). In exams, recognize the triangular-number pattern.

Putting \(n=1\) gives the first term \(4(1)+3=7\). Increasing \(n\) by 1 raises each term by 4, so the sequence is 7, 11, 15, .... Exam tip: check both the first term and common difference.

The governing concept is finding the index of a term from an explicit sequence rule. The displayed terms are successive multiples of 16: 16 × 1, 16 × 2, 16 × 3, and 16 × 4. Hence the nth term is aₙ = 16n. Set this equal to the target value: 16n = 112. Dividing by 16 gives n = 112 ÷ 16 = 7. Therefore, 112 is the seventh term and option C is correct. For comparison, the fifth term is 80, the sixth is 96, and the eighth is 128. These values show why the neighboring position choices are not correct.

For \(a_n=n^2\), substituting \(n=1,2,3,4\) gives 1, 4, 9, 16, which are square numbers. \(n^3\) gives 1, 8, 27 instead. In exams, verify a rule using the first few terms.

The terms can be written as \(5\times1^2,\;5\times2^2,\;5\times3^2,\;5\times4^2\). Hence, the \(n\)th term is \(a_n=5n^2\). The option \(5n\) gives \(5,10,15,20\), so it does not match the sequence. Exam tip: when terms are multiples of square numbers, first test a form involving \(n^2\).

For \(a_n=5n-2\), the consecutive difference is \(a_{n+1}-a_n=5\), which is constant. Hence it is an arithmetic progression. In \(n^2+1\), the difference changes. Exam tip: identify AP terms in the form \(pn+q\).

The successive differences are \(6,10,14\), whose differences are constant at \(4\). Hence, the terms should follow a quadratic expression. Substituting \(n=1,2,3,4\) in \(a_n=2n^2-1\) gives \(1,7,17,31\), so option A is correct. The closest distractor, \(2n^2+1\), gives \(3\) as its first term and is therefore incorrect. Exam tip: test a proposed formula using at least the first three terms.

Putting \(n=1,2,3\) in \(2n-1\) gives 1, 3, and 5, so it generates the odd natural numbers. In contrast, \(2n\) gives 2, 4, and 6, which are even numbers. Exam tip: verify a rule using the first three terms.