Explicit or general rule

Given \(a_n=4n^2-1\), \(a_5=4(5)^2-1=100-1=99\) and \(a_1=4(1)^2-1=3\). Hence, \(a_5+a_1=99+3=102\). Option 100 is only \(4\times 5^2\); it ignores both the \(-1\) and \(a_1\). Exam tip: Substitute each required value of \(n\) separately before adding the terms.

The general term is (a_n=50-5n), and (50-5n=0) gives (n=10). Even in a decreasing sequence the position remains natural.

Substitute n=1, 2, and 3 to obtain the first three terms. a₁=2(1)²+2(1)+1=5, a₂=2(2)²+2(2)+1=13, and a₃=2(3)²+2(3)+1=25. Hence, the correct sequence is (5, 13, 25). Option C has the correct first term, but for n=2 the value is 13, not 11. Exam tip: evaluate the squared term first and substitute each value of n separately.

The numerator is (n^2+1) and the denominator is (n+2), so (a_n=\frac{n^2+1}{n+2}). In a fractional sequence identify the numerator pattern separately.

The direct answer is C: \\(a_n=\\frac{n}{2n+1}\\). Start by numbering the terms: for n=1 the numerator is 1 and the denominator is 3; for n=2 they are 2 and 5; for n=3 they are 3 and 7; for n=4 they are 4 and 9. Thus the numerator is n, while the denominator increases by 2 and is given by \\(2n+1\\). Therefore the general term is \\(a_n=\\frac{n}{2n+1}\\). Option C is correct. Option A gives \\(\\frac{n+1}{2n+1}\\), whose first term is \\(\\frac{2}{3}\\), not \\(\\frac{1}{3}\\). Option B gives \\(\\frac{n}{n+2}\\), whose second term is \\(\\frac{2}{4}\\), not \\(\\frac{2}{5}\\). Option D gives \\(\\frac{2n-1}{n+2}\\), whose first term is \\(\\frac{1}{3}\\) but whose second term is \\(\\frac{3}{4}\\), not \\(\\frac{2}{5}\\). The exam method is to test n=1 and n=2, not just the first term.

The correct rule is \(a_n=3n^2+2\). Substituting \(n=1,2,3,4\) gives \(5,14,29,50\), respectively. The closest distractor, \(a_n=3n^2+n\), gives \(4\) when \(n=1\), so it cannot represent the sequence. Exam tip: verify a proposed general rule using at least the first two or three terms.

Given \(a_n=n^2+cn\), substitute \(n=3\): \(a_3=3^2+3c=9+3c\). Since \(a_3=18\), \(9+3c=18\), so \(3c=9\) and \(c=3\). Therefore, option 3 is correct. If \(c=2\), then \(a_3=9+6=15\), not 18. Exam tip: substitute the specified value of \(n\) directly into the general term.

Given \(a_n=12-2n\), for the first negative term we need \(12-2n<0\). This gives \(n>6\), so the smallest integer value of \(n\) is \(7\). Indeed, \(a_6=0\), which is not negative, whereas \(a_7=12-14=-2\). Therefore, the 7th term is the first negative term. Exam tip: a negative term must be less than \(0\); zero is neither positive nor negative.

The increase over three gaps is (12), so (d=4), hence (a_9=25+3(4)=37). Extend the terms using the common difference.

This is (1^3+1,2^3+1,3^3+1,4^3+1,\ldots), so (a_n=n^3+1). In cube sequences identify the constant shift.

Given \(\frac{n(n+1)}{2}=45\), we get \(n(n+1)=90\), so \(n^2+n-90=0\). Factoring gives \((n-9)(n+10)=0\). Since a term number must be positive, \(n=9\). Therefore, 45 is the 9th term of the sequence. The 10th term is \(\frac{10\times11}{2}=55\), so it is not correct. Exam tip: Clear the denominator first and take the positive root of the resulting quadratic equation.

The magnitude is (6,9,12,15,\ldots) and signs alternate, so (a_n=(-1)^{n+1}3(n+1)). Use the first term sign to decide the power of ((-1)).

For an arithmetic sequence, \(a_n=a_1+(n-1)d\). Here, \(a_4=a_1+3(5)=18\), so \(a_1=3\). Hence \(a_n=3+(n-1)5=5n-2\), making option A correct. In option B, \(a_4=22\), so it cannot be correct. Exam tip: when a term \(a_k\) is given, first find \(a_1=a_k-(k-1)d\), then form the general term.

Given \(a_n=2n^2+n-1\), \(a_6=2(6)^2+6-1=72+6-1=77\) and \(a_2=2(2)^2+2-1=8+2-1=9\). Therefore, \(a_6-a_2=77-9=68\). The value 72 is only \(2(6)^2\), so it ignores the \(+6-1\) part of the rule. Exam tip: calculate each required term separately before finding their difference.

The first term is (2) and the common difference is (7), so (a_n=2+(n-1)7=7n-5). In a linear sequence the coefficient of (n) is the common difference.

The first term is (31) and the difference is (-5), so (a_n=31+(n-1)(-5)=36-5n). In a decreasing sequence take the common difference as negative.

Given \(a_n=3n^2+2n\), substitute \(n=4\) to find the fourth term: \(a_4=3(4)^2+2(4)=3\times16+8=48+8=56\). Therefore, 56 is correct. The option 54 may result from an error while squaring or multiplying. Exam tip: In expressions with powers, evaluate \(n^2\) first, then multiply and add.

The numerator is (2n+1) and the denominator is (3n+1), so (a_n=\frac{2n+1}{3n+1}). In fractions identify the numerator and denominator rules separately.

The governing concept is substitution into an explicit or general rule. Since the formula directly gives the term at position n, substitute n = 1, 2, 3, and 4 in order. For n = 1, a₁ = 2(1)(1 + 1) = 4. For n = 2, a₂ = 2(2)(2 + 1) = 12. For n = 3, a₃ = 2(3)(3 + 1) = 24. For n = 4, a₄ = 2(4)(4 + 1) = 40. Thus the first four terms are 4, 12, 24, 40, so option B is correct. Option A does not even have the correct first term. Options C and D likewise fail when n = 1 is substituted, showing that their values come from shifting or misusing the index.

For the fourth term, substitute \(n=4\): \(a_4=7\cdot3^{4-1}=7\cdot3^3=7\cdot27=189\). Therefore, 189 is correct. The value 63 may result from incorrectly using \(7\cdot3^2\). Exam tip: In a general-term formula, substitute the term number first and simplify the exponent carefully.

This is (4^2,5^2,6^2,7^2,\ldots), so (a_n=(n+3)^2). In square sequences relate the base number to (n).

(a_5=13) and (b_3=19), so the sum is (32). With two rules the term numbers may differ, so use them carefully.

The rule (a_n=2n^2+3n+3) gives (a_9=192), so the shown options do not match. This is an option-consistency check question.

Substituting \(n=1,2,3,4\) into \(a_n=n^2+3n\) gives \(4,10,18,28\), respectively. Therefore, option A is correct. Option C gives the first two terms as \(4,10\), but its third term is \(20\), not \(18\). Exam tip: Verify a proposed sequence rule using at least three or four terms.

This is an arithmetic progression with first term 11 and common difference 7. Its general term is \(a_n=11+(n-1)\times7=7n+4\). Setting \(7n+4=74\) gives \(n=10\), so 74 is the tenth term. The ninth term is \(7(9)+4=67\), not 74. Exam tip: To find the position of a number in an AP, equate it to \(a_n\) and solve for \(n\).