Explicit or general rule

The governing concept is an explicit or general rule, which expresses a term directly in terms of its position n rather than using the preceding term. The sequence 3, 6, 9, 12 consists of successive multiples of 3. Testing aₙ = 3n gives a₁ = 3(1) = 3, a₂ = 3(2) = 6, a₃ = 3(3) = 9 and a₄ = 3(4) = 12, so it matches every displayed term. Therefore option C is correct. The rule n + 3 gives 4 as its first term, 2n gives 2 as its first term, and 3n + 1 gives 4 as its first term. Since each of these already fails at n = 1, none can represent the sequence.

The governing concept is an explicit rule, where the nth term is obtained directly from n. The displayed sequence is made of consecutive multiples of 5: 5×1, 5×2, 5×3 and 5×4. Thus the natural rule is aₙ = 5n. Checking it gives a₁ = 5, a₂ = 10, a₃ = 15 and a₄ = 20, exactly as required. Therefore option A is correct. The rule n + 5 produces 6 as the first term, 4n produces 4 as the first term, and 5n − 1 produces 4 as the first term; each fails immediately at n = 1. The constant difference is 5, but that fact must still be combined with the correct first-term position to obtain 5n.

The direct answer is Option B: aₙ=2n. The sequence is 2, 4, 6, 8, and so on. These are consecutive positive even numbers. For a general term, use the position n: at n=1, 2n=2; at n=2, 2n=4; at n=3, 2n=6; at n=4, 2n=8. Since the rule produces every listed term, aₙ=2n is correct. Option A, aₙ=n, produces 1, 2, 3, 4, so it is the sequence of counting numbers, not this sequence. Option C, aₙ=2n+1, produces 3, 5, 7, 9, which are odd numbers and start at 3. Option D, aₙ=n+2, produces 3, 4, 5, 6, so it does not match the first term or the constant pattern correctly. The subscript n means the position of the term, beginning with n=1. Multiplying the position by 2 gives the even number at that position. Exam cue: the nth positive even number is 2n; the nth odd number is 2n−1.

The direct answer is C: a_n=2n-1. An explicit rule gives a term directly from its position n. The sequence 1,3,5,7 increases by 2 each time and consists of positive odd numbers. Test the rule: for n=1, 2(1)-1=1; for n=2, 2(2)-1=3; for n=3, 2(3)-1=5; for n=4, 2(4)-1=7. Thus it matches every displayed term. Option A, a_n=2n, gives 2,4,6,8, so it starts too high and produces even numbers. Option B, n+1, gives 2,3,4,5, not the sequence. Option C is correct. Option D, 3n-2, gives 1,4,7,10, whose differences are 3 rather than 2. Memory cue: the nth odd number is 2n-1.

For the first three terms, substitute \(n=1,2,3\). This gives \(a_1=1^2=1\), \(a_2=2^2=4\), and \(a_3=3^2=9\). Therefore, the sequence begins as \((1,4,9)\). Option A lists the values of \(n\), not their squares. Exam tip: Substitute \(n=1,2,3\) into the general term to find the initial terms quickly.

The governing concept is the explicit rule for an arithmetic sequence. The common difference is d = 7 − 4 = 3. For an arithmetic sequence, aₙ = a₁ + (n − 1)d. Substituting a₁ = 4 and d = 3 gives aₙ = 4 + 3(n − 1) = 3n + 1. Therefore option A is correct. Checking n = 1, 2, and 3 gives 4, 7, and 10, exactly the listed terms. Option B produces 4, 8, 12, so its difference is too large. Option C starts with 2 rather than 4, and option D produces 4, 5, 6, so its difference is too small. Thus only option A satisfies both the first term and the constant difference.

For \(a_n=3n-2\), \(a_{n+1}-a_n=3\), which is constant; hence it is an arithmetic progression. In \(n^2+1\), the differences vary. Exam tip: check the first difference.

This question tests recognition of an explicit rule. Every displayed term is obtained by multiplying its position number by 10: the first term is 10 × 1, the second is 10 × 2, the third is 10 × 3, and the fourth is 10 × 4. Therefore the general term is a_n = 10n, so option C is correct. It can also be verified using the arithmetic-sequence formula: a_1 = 10 and d = 10, hence a_n = 10 + (n − 1)10 = 10n. Option A gives 11 as its second term, option B gives 5, 10, 15, …, and option D gives 9 for n = 1. Thus none of those alternatives reproduces all the listed terms.

In \(a_n=2n+1\), any term is obtained directly by substituting its position \(n\), so it is explicit. Options A and D depend on a previous term and are recursive rules. Exam tip: an explicit rule usually expresses \(a_n\) directly in terms of \(n\).

Direct answer: A, \\(a_n=5n+1\\). This is an arithmetic sequence because each term increases by the same amount: 11-6=5, 16-11=5 and 21-16=5. For an arithmetic sequence, the nth term is first term plus \\(n-1\\) times the common difference: \\(a_n=6+(n-1)5=6+5n-5=5n+1\\). Check: for \\(n=1\\), the result is 6; for \\(n=2\\), it is 11; for \\(n=3\\), it is 16. Option A matches every check. Option B gives 6, 12, 18, so its difference is 6 and it fails at the second term. Option C gives 4 for \\(n=1\\), not 6. Option D gives 6, 7, 8, so its difference is 1. Memory cue: for an arithmetic sequence, use first term plus \\(n-1\\) times the common difference.

The direct answer is Option B: 10. An explicit rule gives a term directly from its position. The first term is the term whose position is n=1, so substitute 1 into aₙ=n+9: a₁=1+9=10. Therefore Option B is correct. Option A, 9, would result from using n=0, but in this school sequence the first term is normally indexed by n=1 unless the question explicitly states zero-based indexing. Option C, 11, comes from using n=2, so it is the second term, not the first. Option D, 12, comes from using n=3, so it is the third term. The key distinction is between the formula and the position: n is not an unknown to guess; it is the term number requested. To find the first term, always set n=1. The first few terms confirm the pattern: a₁=10, a₂=11, a₃=12. Exam cue: “first term” means substitute 1, not 0, unless the indexing convention is clearly changed.

This is an arithmetic progression with first term \(a=9\) and common difference \(d=9\). Thus, \(a_n=a+(n-1)d=9+(n-1)\times9=9n\). Therefore, the correct rule is \(a_n=9n\). The rule \(a_n=9n+1\) gives 10 as the first term, so it is incorrect. Exam tip: substitute \(n=1\) to check whether a proposed rule gives the first term correctly.

In an AP, the difference between consecutive terms is constant. For \(a_n=5n-1\), \(a_{n+1}-a_n=5\), so it is an AP. In \(n^2+1\), the differences change. Exam tip: a rule of the form \(pn+q\) usually represents an AP.

Each term is obtained by multiplying by (7), so (a_n=7n). In exams, identify table-based sequences quickly.

The direct answer is A: (7,9). The explicit formula is a_n=2n+5. To find a_1, put n=1: a_1=2(1)+5=7. To find a_2, put n=2: a_2=2(2)+5=9. Therefore the ordered pair is (7,9). Option A is correct because both substitutions are accurate. Option B, (6,8), would come from using a different formula or subtracting 1, so it does not follow the given rule. Option C, (5,7), ignores the 2n part and gives values that are too small. Option D, (8,10), adds one too many to both correct results. A common mistake is to use n=0 for the first term; in the usual sequence notation here, a_1 means n=1. Memory cue: replace n by the requested position, then simplify.

The governing concept is the explicit nth-term rule for an arithmetic sequence. The difference between consecutive terms is constant: 15 − 12 = 3, 18 − 15 = 3, and 21 − 18 = 3. Using a_n = a_1 + (n − 1)d, with first term a_1 = 12 and common difference d = 3, we obtain a_n = 12 + 3(n − 1) = 3n + 9. Therefore option C is correct. Substitution confirms the rule: n = 1 gives 12, n = 2 gives 15, n = 3 gives 18, and n = 4 gives 21. Option A begins with 15, option B has common difference 12, and option D has common difference 4. Thus only C matches both the first term and the constant change.

Since \(a_n=4n+3\) has the linear form \(pn+q\), it represents an arithmetic progression with common difference \(p=4\). Consecutive terms differ by 4; their ratio is not constant. Exam tip: the coefficient of \(n\) gives the common difference.

The governing pattern is the sequence of consecutive square numbers. The terms can be written as 1^2, 2^2, 3^2, and 4^2. Since the term in position n is the square of n, its explicit rule is a_n = n^2, making option C correct. Substitution verifies the rule: for n = 1, a_1 = 1; for n = 2, a_2 = 4; for n = 3, a_3 = 9; and for n = 4, a_4 = 16. Option A produces 2, 3, 4, 5, option B produces even numbers, and option D produces multiples of 3. Although the differences between the terms are 3, 5, and 7 rather than constant, the square-number rule identifies the sequence directly.

The general term is
a_n=n^2+1
. Substituting
n=4
gives
a_4=4^2+1=16+1=17
. Therefore, 17 is correct. Option 16 is only the value of
4^2
and misses the added 1. Exam tip: Substitute the term number carefully for
n
in the general rule.

For \(n=1\), \(a_1=1^2+1=2\); for \(n=2\), \(a_2=2^2+1=5\); and for \(n=3\), \(a_3=3^2+1=10\). Hence, the correct general term is \(a_n=n^2+1\). The close option \(a_n=n^2\) gives the first term as 1, not 2. In an exam, substitute \(n=1,2,3\) to verify a proposed general term quickly.

At (n=1), it gives (9), and at (n=2), it gives (8). In exams, for a decreasing difference of (1), think (a_n=\text{constant}-n).

Given \(a_n=15-2n\), substitute \(n=5\): \(a_5=15-2(5)=15-10=5\). Therefore, 5 is correct. The value 7 would result from incorrectly subtracting \(n\) instead of \(2n\). Exam tip: To find a specified term from a general rule, substitute the given value of \(n\) carefully.

The direct answer is B: a_n=15-2n. The sequence starts at 13 and decreases by 2 each time, so its common difference is -2. For a sequence with first term a_1 and constant difference d, the explicit rule is a_n=a_1+(n-1)d. Here a_n=13+(n-1)(-2)=13-2n+2=15-2n. Check it: n=1 gives 13, n=2 gives 11, n=3 gives 9, and n=4 gives 7. Option A, 2n+11, increases and starts at 13, so it gives 15 next. Option B matches every term. Option C, 13-2n, gives 11 as the first term because it forgets the n-1 adjustment. Option D, 11+n, increases by 1. Memory cue: a decreasing sequence has a negative difference.

The general rule is \(a_n=2^n\). For the third term, substitute \(n=3\): \(a_3=2^3=8\). Therefore, 8 is the correct option. The value 4 would result from an exponent of 2, not 3. Exam tip: substitute the required term number carefully for \(n\) in the general-term formula.

Each term is a power of (2), so (a_n=2^n). In exams, recognize geometric-type growth using powers.