Explicit or general rule

The increase over six gaps is (54), so (d=9), hence (a_{15}=83+4(9)=119). Extend the terms using the common difference.

With only four given terms, the general term of a sequence is not uniquely determined. For example, the polynomial in option C gives \(1,15,63,195\) for \(n=1,2,3,4\), but many other rules can also produce these same four terms. Option A fits only the first two terms; at \(n=3\), it gives \(53\), not \(63\). Exam tip: Test a proposed rule against every listed term and check whether enough information is available to define a unique pattern.

When \(n\) increases by 1 in \(a_n=4n-1\), the term increases by 4, so \(a_{n+1}-a_n=4\) is constant. Hence it is an arithmetic progression, not a GP, which needs a constant ratio. Exam tip: in \(pn+q\), the common difference is \(p\).

For \(a_n=5n^2+2n\), substituting \(n=1,2,3,4\) gives \(7,24,51,88\), respectively. Hence, it is the required explicit rule. Although \(a_n=3n^2+4n\) gives the first term as 7, it gives 20, not 24, when \(n=2\). Exam tip: test a rule using at least the first three terms.

The consecutive differences are 5−2=3, 8−5=3, and 11−8=3, so the common difference is constant and the sequence is an AP. In option B, the differences 1, 2, 4 change. Exam tip: check consecutive differences to identify an AP.

The successive differences are \(9,15,21\), and their second differences are constant: \(6,6\). Hence the rule is quadratic, with coefficient of \(n^2\) equal to \(6/2=3\). Substituting in \(a_n=3n^2-1\) gives \(a_1=2, a_2=11, a_3=26\), and \(a_4=47\). Option D is linear and would require equal first differences. Exam tip: constant second differences usually indicate a quadratic sequence rule.

Substitute \(n=5\): \(a_5=\frac{5(4\times5+1)}{2}=\frac{5(21)}{2}=\frac{105}{2}\). Hence, option C is correct. \(\frac{101}{2}\) may result from evaluating \(4\times5+1\) incorrectly. Exam tip: after substituting the term number in a general term, follow the order of operations carefully.

The governing concept is an explicit rule: a formula must produce the term directly from its position n. Test option A at the first four positive integer values. For n = 1, a₁ = 1(4 + 1)/2 = 5/2. For n = 2, a₂ = 2(8 + 1)/2 = 9. For n = 3, a₃ = 3(12 + 1)/2 = 39/2, and for n = 4, a₄ = 4(16 + 1)/2 = 34. Thus option A reproduces every listed term. Option B gives 5/2 for the first term but fails at n = 2, option C gives 3, and option D gives 2 at n = 1, so they cannot be correct.

Given \(a_n=(3n-2)^2\). Substituting \(n=4\), \(a_4=(3\times4-2)^2=(12-2)^2=10^2=100\). Therefore, 100 is the correct option. \(121\) would result only if the expression inside the square were 11, which it is not here. Exam tip: substitute the value of \(n\) first, then simplify the bracket before applying the exponent.

The terms are \(1^2,4^2,7^2,10^2\). Their square roots, \(1,4,7,10\), form an arithmetic sequence whose \(n\)th term is \(3n-2\). Hence the explicit rule is \(a_n=(3n-2)^2\). In option B, substituting \(n=2\) gives 10, not 16. Exam tip: for sequences of perfect squares, first check the pattern in their square roots.

Given \(a_n=150-11n\), substitute \(n=7\): \(a_7=150-11\times7=150-77=73\). Hence, \(73\) is correct. \(77\) is only the value of \(11\times7\), not the value of \(a_7\). Exam tip: substitute the term number carefully, multiply first, and then subtract.

At (n=1) it gives (139), and at (n=2) it gives (128), so (a_n=150-11n). In exams, match the first term of a decreasing sequence.

The governing concept is using an explicit sequence rule to identify the position of a specified term. Set the nth-term expression equal to the given value: 8n − 5 = 155. Adding 5 to both sides gives 8n = 160, and dividing by 8 gives n = 20. A check confirms that a₂₀ = 8(20) − 5 = 160 − 5 = 155. Hence option C is correct. Option A would produce 139, option B would produce 147, and option D would produce 163. These distractors result from choosing a nearby position or making an arithmetic error, but none satisfies the defining equation.

Its rule is (a_n=11n-7), and (11n-7=125) gives (n=12). In exams, equate the given term to the general term.

The term number starts with \(n=1\). Substituting \(n=1,2,3,4\) into \(a_n=2^n+n^2\) gives \(3,8,17,32\), respectively. The close distractor \(2n^2+1\) gives \(3,9\) as its first two terms, so it does not fit the sequence. Exam tip: verify a proposed general term using at least the first three values of \(n\).

Putting n=3, we get a_3=3^3+3^2-1=27+9-1=35. Therefore, the correct answer is 35. A value such as 33 results from an incorrect calculation of the final subtraction; 27+9-1 equals 35. Exam tip: when finding a term from an explicit rule, substitute the given value of n in every occurrence of n, including powers and squares.

The correct rule is \(a_n=3^n+n^2-1\). Substituting \(n=1,2,3,4\) gives \(3,12,35,96\), respectively. Although \(a_n=3n^2\) gives the first two terms as 3 and 12, it gives 27 as the third term instead of 35, so it is not the rule for the sequence. In exams, test a proposed general term with at least the first three terms.

Substitute \(n=5\): \(a_5=7\cdot2^{5-1}+4=7\cdot2^4+4=7\cdot16+4=116\). Therefore, the correct answer is 116. The value 112 is only \(7\cdot16\); it misses the final \(+4\). Exam tip: evaluate \(n-1\) first, then calculate the power.

An explicit rule must generate each term from n without needing the preceding term. Test option A: at n = 1, it gives 7·2⁰ + 4 = 11; at n = 2, it gives 7·2¹ + 4 = 18; at n = 3, it gives 7·2² + 4 = 32; and at n = 4, it gives 7·2³ + 4 = 60. Therefore A exactly matches the sequence. Option B gives 11, 22, 33, 44 and is not the sequence. Option C begins 11 but gives 13 at n = 2, while option D gives 11 at n = 1 but 23 at n = 2. Matching several positions, rather than only the first term, establishes the answer.

Given \(a_n=4n^2+5n-6\), \(a_3=4(3)^2+5(3)-6=45\) and \(a_5=4(5)^2+5(5)-6=119\). Therefore, \(a_3+a_5=45+119=164\). A nearby value such as 162 can result from a small addition or multiplication error. Exam tip: substitute each value of \(n\) separately before adding the terms.