Explicit or general rule

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.

Putting (n=1) and (n=2) gives (7) and (9). In exams, find the first two terms to check the rule.

The relevant concept is the nth-term rule for an arithmetic sequence. The common difference is d = 15 − 12 = 3. Using a_n = a_1 + (n − 1)d with a_1 = 12 gives a_n = 12 + 3(n − 1) = 12 + 3n − 3 = 3n + 9. Hence option C is correct. A direct check gives a_1 = 3(1) + 9 = 12, a_2 = 15, a_3 = 18, and a_4 = 21. Option A gives 15 when n = 1, so it starts incorrectly; option B gives 12, 24, 36, …, and option D gives 12, 16, 20, …, which has a different common difference. Therefore only option C fits the entire pattern.

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 first term is (13) and the difference is (-2), so (a_n=15-2n). In exams, treat a decreasing difference as negative.

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.

The general rule gives each term by substituting its position for \(n\). Putting \(n=2\), we get \(a_2=3^2=9\). Option 8 is incorrect because it equals \(2^3\), whereas the base here is 3 and the exponent is 2. Exam tip: first substitute the term number for \(n\), then evaluate the power.

This is a geometric sequence in which each term is 3 times the previous term. Starting with \(n=1\), we get \(a_1=3^1=3\), \(a_2=3^2=9\), \(a_3=3^3=27\), and \(a_4=3^4=81\). Hence, the correct general term is \(a_n=3^n\). The rule \(a_n=3n\) would give 3, 6, 9, 12, so it is not correct. Exam tip: Substitute \(n=1\) in a proposed rule to check the first term.

Given \(a_n=n^2-n\), substitute \(n=5\): \(a_5=5^2-5=25-5=20\). Hence, \(20\) is correct. \(25\) is only the value of \(5^2\); subtracting \(5\) is also required. Exam tip: substitute the value of \(n\) in every occurrence in the formula.

For term numbers \(n=1,2,3,4\), the rule \(n^2-n\) gives \(0,2,6,12\), respectively. Therefore, the correct general term is \(a_n=n^2-n\). The rule \(a_n=2n\) gives 2 as its first term, so it does not match the sequence. Exam tip: verify a general term by substituting \(n=1\) and \(n=2\) and checking the first two terms.

Since \(a_n=7\), every term remains 7 regardless of \(n\), so it is a constant sequence. Its common difference is 0, not 7. Exam tip: list terms as 7, 7, 7 to identify it quickly.

The sequence follows an explicit arithmetic rule. Its common difference is 11 − 7 = 4, so the coefficient of n in the general term is 4. Applying a_n = a_1 + (n − 1)d gives a_n = 7 + (n − 1)4 = 7 + 4n − 4 = 4n + 3. Therefore option A is correct. Substitution confirms that n = 1 gives 7, n = 2 gives 11, n = 3 gives 15, and n = 4 gives 19. Option B gives 7, 14, 21, …; option C starts with 1; and option D gives 7, 8, 9, …. These alternatives either use the wrong difference or fail to preserve the pattern, so they cannot be the required general rule.

In \(a_n=5n+2\), the coefficient of \(n\) is 5, so each successive term increases by 5. Hence it is an AP, not a GP. Exam tip: in the form \(pn+q\), the common difference is \(p\).