nth term

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. The formula aₙ = a₁ + (n − 1)d therefore gives aₙ = 10 + (n − 1)3 = 10 + 3n − 3 = 3n + 7. Option A is correct. Verification is straightforward: when n = 1, the formula gives 3 + 7 = 10; when n = 2, it gives 6 + 7 = 13; and when n = 4, it gives 12 + 7 = 19. Option B has an incorrect coefficient, option C gives 4 for n = 1, and option D produces a difference of 1 rather than 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 nth-term rule. The sequence is defined by aₙ = n(n + 1)/2, and the requested index is n = 6. Substitute this value carefully: a₆ = 6(6 + 1)/2 = 6×7/2 = 42/2 = 21. Therefore, option C is correct. The distractor 15 is the value obtained for n = 5, because 5×6/2 = 15, not for the sixth term. The values 18 and 24 do not result from correct substitution into the given formula. The important distinction is that the formula already specifies the value of each indexed term, so there is no need to add six to a previous term or replace the expression 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 determining the index of a value from an explicit sequence rule. The terms are consecutive multiples of 16: 16×1, 16×2, 16×3, and 16×4. Hence the nth term is aₙ = 16n. To locate 112, set 16n = 112 and divide by 16, obtaining n = 7. Therefore, 112 is the seventh term, so option C is correct. Checking nearby terms gives the fifth term as 16×5 = 80, the sixth as 16×6 = 96, the seventh as 16×7 = 112, and the eighth as 16×8 = 128. Since only the seventh multiple equals 112, the other choices are excluded.

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.

For \(a_n=3n^2+2\), substituting \(n=1,2,3,4\) gives \(5,14,29,50\), respectively. Hence, it generates the given sequence. Although \(a_n=n^2+4\) gives the first term as 5, its second term is 8, not 14. Exam tip: test an nth-term formula using at least the first two or three terms.

For the fourth term, substitute \(n=4\): \(a_4=\frac{4(4+3)}{2}=\frac{4\times7}{2}=14\). Therefore, the correct answer is 14. A value such as 16 can result from incorrectly evaluating \(4+3\). Exam tip: in an nth-term formula, substitute the given value of \(n\) first and then simplify step by step.

The differences (3,4,5) are increasing, so the next difference is (6) and the fifth term is (20). In exams, observe changing differences in order.

The governing concept is substitution into an explicit nth-term rule. The formula gives every term directly from its index, so evaluate it separately at n = 5 and n = 6. For n = 5, a₅ = 5² + 5 + 2 = 25 + 5 + 2 = 32. For n = 6, a₆ = 6² + 6 + 2 = 36 + 6 + 2 = 44. Therefore a₅ + a₆ = 32 + 44 = 76, so option B is correct. An efficient check is to combine the expressions: (25 + 5 + 2) + (36 + 6 + 2) = 61 + 15 = 76. The value 78 may result from an addition mistake, while 80 or 74 can arise from incorrectly substituting one index or mishandling a square. The supplied original key should therefore be corrected to B.

The governing concept is direct substitution into an explicit sequence rule, together with checking whether the index is even or odd. Put n = 6 into aₙ = (−1)ⁿ(2n + 3): a₆ = (−1)⁶(2 × 6 + 3). Because 6 is even, (−1)⁶ = 1. The remaining expression is 12 + 3 = 15, so a₆ = 1 × 15 = 15. Therefore option A is correct. Option B, −15, would occur if the sign factor were negative, as it is for an odd index. Options C and D are not obtained because 2n + 3 equals 15, not 12, when n = 6. Evaluating the sign factor and the bracket separately prevents both common errors.

The governing concept is applying one explicit sequence rule at two different indices and then subtracting. First calculate the term three places after aₙ: aₙ₊₃ = 3(n + 3) + 2 = 3n + 9 + 2 = 3n + 11. Now subtract the original term: aₙ₊₃ − aₙ = (3n + 11) − (3n + 2) = 9. Therefore option A is correct. The same result follows from the constant rate of change: every increase of 1 in the index increases the term by 3, so an increase of 3 in the index increases the term by 3 × 3 = 9. Option C counts only one step, and option B does not use the three-step change correctly. Option D is part of aₙ₊₃, not the requested difference.

The governing concept is evaluation of an explicit nth-term rule. Since \(a_n=6n-5\), substitute the requested index \(n=18\) directly into the formula: \(a_{18}=6(18)-5=108-5=103\). Therefore option B is correct. No recursive calculation or summation is needed because the formula already gives any term directly from its position. Option A would result from subtracting 7 instead of 5, option C from an arithmetic error after multiplication, and option D from subtracting 1 rather than 5. The sequence has common difference 6, which is consistent with this linear rule.

The governing concept is direct substitution into an explicit nth-term formula. Every occurrence of n must be replaced by 7. Thus a₇ = [3(7) − 2]/[2(7) + 1] = (21 − 2)/(14 + 1) = 19/15. The numerator is 19 and the denominator is 15; because 19 is prime and does not divide 15, the fraction is already in simplest form. Therefore option C is correct. A common error is to substitute 7 in only one part of the expression, to calculate 2(7) + 1 incorrectly, or to simplify the fraction improperly. Options A, B, and D represent such arithmetic or substitution mistakes rather than valid evaluations of the given rule.