Explicit or general rule

The governing concept is the explicit formula of an arithmetic sequence. The common difference is constant: 16−9=7, 23−16=7, and 30−23=7. Thus a_n=a_1+(n−1)d=9+(n−1)7=9+7n−7=7n+2. Option D is therefore correct. Substitution verifies the rule: for n=1, a_1=7(1)+2=9; for n=2 it is 16; for n=3 it is 23; and for n=4 it is 30. Option A has the wrong difference and first term, option B gives 5 when n=1, and option C gives 9 initially but then increases by only 2. Checking both the first term and common difference prevents these errors.

Given \(a_n=60-5n\), substitute \(n=9\) to find the ninth term: \(a_9=60-5(9)=60-45=15\). Therefore, 15 is correct. The value 20 could result from incorrectly calculating \(5\times9\) as 40. Exam tip: substitute the term number for \(n\) carefully in a general-term formula.

At (n=1) it gives (47) and at (n=2) it gives (43) so (a_n=51-4n). In exams match the first term in a decreasing sequence.

The governing concept is using an explicit rule to identify the position of a specified term. We are given a_n=5n+4 and want a_n=64, so set the formula equal to 64: 5n+4=64. Subtracting 4 from both sides gives 5n=60, and dividing by 5 gives n=12. Therefore, option A is correct: the twelfth term is 64. Substitution confirms it because a_12=5(12)+4=60+4=64. The other choices fail the formula: n=11 gives 59, n=13 gives 69, and n=10 gives 54. The essential method is to equate the explicit expression with the required value and solve for the positive integer term number.

Its rule is (a_n=3n-1) and (3n-1=38) gives (n=13). In exams form the general term first to find the term number.

These terms are (3^2,4^2,5^2,6^2) so (a_n=(n+2)^2). In exams recognize shifted square numbers.

Given \(a_n=(n+2)^2\). Substituting \(n=5\), we get \(a_5=(5+2)^2=7^2=49\). Therefore, 49 is correct. The value 36 would result from using 6 before squaring, which is not the correct substitution here. Exam tip: substitute the value of \(n\) first, simplify the bracket, and then square it.

Given \(a_n=n^2+2n\), substitute \(n=6\): \(a_6=6^2+2(6)=36+12=48\). Therefore, the correct answer is \(48\). A value such as \(46\) may result from not evaluating or adding the squared term and the \(2n\) term correctly. Exam tip: substitute the term number for every occurrence of \(n\) in the expression.

Its rule is (a_n=n^2) and (n^2=81) gives (n=9). In exams identify the term number of square numbers.

Given \(a_n=\frac{n(n+1)}{2}+2\), substitute \(n=6\): \(a_6=\frac{6(6+1)}{2}+2=\frac{42}{2}+2=21+2=23\). Hence, the correct answer is 23. The value 21 is only \(\frac{6\times7}{2}\); the given \(+2\) must still be added. Exam tip: when evaluating a general term, substitute the value of \(n\) carefully and perform the operations in order.

(\frac{n(n+1)}{2}+2) gives (3,5,8,12). In exams think of triangular numbers when differences are (2,3,4).

The first differences are \(5,7,9\), increasing by 2 each time, so the general term should be quadratic. Substituting \(n=1,2,3,4\) in \(a_n=n^2+n-2\) gives \(0,5,12,21\), respectively. Option A gives \(3\) as the second term, so it does not fit. Exam tip: verify a proposed general term by checking at least the first four values of \(n\).

Set \(n^2+n-2=70\). This gives \(n^2+n-72=0\), or \((n+9)(n-8)=0\). The roots are \(n=8\) and \(n=-9\), but a term number must be positive, so \(n=8\) is correct. For example, \(n=7\) gives 54, not 70. Exam tip: reject a negative root when it cannot represent a term number.

The correct rule is \(a_n=2^n+2\). Substituting \(n=1,2,3,4\) gives \(4,6,10,18\), respectively. Although \(a_n=2n+2\) gives the first two terms as 4 and 6, its third term is 8, not 10. Exam tip: verify an explicit rule using at least the first three terms.

The general term is \(a_n=2^n+2\). Substituting \(n=6\), we get \(a_6=2^6+2=64+2=66\). Although 64 is a close distractor, it is only the value of \(2^6\); the given \(+2\) must also be added. Exam tip: substitute the term number for \(n\), evaluate the power first, and then perform the remaining operations.

Substituting \(n=1,2,3,4\) in \(a_n=3^n+1\) gives \(4,10,28,82\), respectively. Hence, the correct general term is \(a_n=3^n+1\). Although \(a_n=n^3+3\) gives the first term as 4, its second term is 11, not 10. Exam tip: verify a proposed general term using at least the first two or three terms.

Given \(a_n=3^n+1\), substitute \(n=4\): \(a_4=3^4+1=81+1=82\). Option 81 is only the value of \(3^4\); it misses the added 1 in the rule. Exam tip: after substituting the index in a general term, include both the power and the constant term.

(6\cdot2^n) gives (12,24,48,96). In exams check both the base and multiplier in geometric growth.

Given \(a_n=3\cdot2^{n-1}\). Substituting \(n=7\), \(a_7=3\cdot2^{7-1}=3\cdot2^6=3\cdot64=192\). The value 96 can result from incorrectly using \(2^5\). Exam tip: substitute the term number into the exponent \(n-1\) before calculating.

The first term is (3) and each term is multiplied by (2) so (a_n=3\cdot2^{n-1}). In exams check both the first term and ratio.