nth term

This is a cube sequence and (512=8^3), so it is the (8)th term. In a cube sequence, the term number matches the cube root.

The index increases by (4) and the common difference is (6), so the difference is (24). In a linear rule, the coefficient is the common difference.

The correct answer is option A: \\(a_n=6\cdot4^{n-1}\\). This is a geometric sequence because each term is obtained by multiplying the previous term by 4: \\(24/6=4\\), \\(96/24=4\\), and \\(384/96=4\\). For a geometric sequence, the nth term is \\(a_n=ar^{n-1}\\), where \\(a\\) is the first term and \\(r\\) is the common ratio. Here \\(a=6\\) and \\(r=4\\), so \\(a_n=6\cdot4^{n-1}\\). Option A gives the first term when \\(n=1\\): \\(6\cdot4^0=6\\), so it works. Option B gives 4 when \\(n=1\\), not 6. Option C gives \\(24\\) for \\(n=1\\), so its exponent is one step too large. Option D also gives 24 for the first term, not 6. Memory cue: first term stays outside, and the exponent is one less than the term number.

Given \(a_n=3n^2+2n-5\), we get \(a_{n+2}=3(n+2)^2+2(n+2)-5=3n^2+14n+11\). Hence, \(a_{n+2}-a_n=(3n^2+14n+11)-(3n^2+2n-5)=12n+16\). The option \(12n+20\) results from an incorrect subtraction of the constant terms. Exam tip: substitute \(n+2\) into every term first, then combine like terms carefully.

From (9d=45), (d=5), so (a_{20}=59+8\cdot5=99). It is easier to find a far term from a nearby given term.

The terms should be matched carefully with the option. Test all starting terms before finalizing the rule.

Given \(a_n=4n-7\), substitute the complete index \(2n+1\) for \(n\): \(a_{2n+1}=4(2n+1)-7=8n+4-7=8n-3\). Hence, the correct answer is \(8n-3\). The option \(8n-7\) results from incorrectly ignoring the \(+4\) produced by the \(+1\) in the index. Exam tip: always place a compound index in brackets before substituting it into a formula.

This is an arithmetic progression with first term 14 and common difference 8. Its nth term is \(a_n=14+(n-1)\times 8\). On setting \(14+(n-1)\times 8=126\), we get \(n-1=14\), hence \(n=15\). Therefore, 126 is the 15th term. The 14th term is 118, so it is a close but incorrect option. Exam tip: while finding a term number, be careful to use \(n-1\) in the AP formula.

Substitute \(n=9\) in \(a_n=\frac{n(n+3)}{2}\): \(a_9=\frac{9(9+3)}{2}=\frac{9\times12}{2}=54\). Therefore, 54 is correct. A value such as 48 may result from an error in evaluating \(9+3\) or in multiplication. Exam tip: after substituting \(n\), simplify the expression inside brackets first.

The successive differences are \(7,11,15,19\), and their second differences are constant at \(4\). Hence the sequence has a quadratic nth-term rule. With \(a_n=2n^2+n\), we get \(a_1=3\), \(a_2=10\), \(a_3=21\), and \(a_4=36\). The close distractor \(2n^2+n-1\) gives the first term as \(2\), not \(3\). Exam tip: a constant second difference usually indicates an \(n^2\)-based rule.

Putting \(n=6\), \(a_6=5\cdot2^{6-1}+1=5\cdot2^5+1=5\cdot32+1=161\). Therefore, 161 is correct. Option 160 would result from incorrectly omitting the final \(+1\). Exam tip: Substitute the value of \(n\) into the exponent \(n-1\) first, then calculate step by step.

For odd (n), terms are negative and for even (n), terms are positive, so (a_n=2(-1)^n n). In alternating signs, test first at (n=1).

To find the position of 87, set \(a_n=87\): \(n^2+6=87\). Thus, \(n^2=81\), giving \(n=9\), since a term number must be a positive integer. Hence, 87 is the 9th term of the sequence. The 8th term is \(8^2+6=70\), so it is not correct. Exam tip: To find which term has a given value, equate \(a_n\) to that value and solve for \(n\).

Each term is halved, so \(a_n=18\left(\frac{1}{2}\right)^{n-1}\). In a geometric sequence, the exponent starts with (n-1).

Given \(a_n=3n^2-7n+6\), \(a_2=3(2)^2-7(2)+6=12-14+6=4\), while \(a_9=3(9)^2-7(9)+6=243-63+6=186\). Therefore, the difference is \(186-4=182\). A value such as 180 can result from a small calculation or subtraction error. Exam tip: substitute each value of \(n\) separately before finding the difference.

The first term is (12) and the difference is (13), so (a_{30}=12+29\cdot13=389). The (30)th term includes (29) differences.

Given \(a_n=2n^2+5n\), \(a_6=2(6)^2+5(6)=72+30=102\) and \(a_3=2(3)^2+5(3)=18+15=33\). Therefore, \(a_6-a_3=102-33=69\). The value 66 can result from an error while calculating \(a_3\) or subtracting. Exam tip: evaluate each term separately before finding their difference.

The terms are (5\cdot1^2,5\cdot2^2,5\cdot3^2,\ldots), so (a_n=5n^2). Recognize the square pattern with a coefficient.

Given \(a_n=25-6n\) and \(a_n=1\), set \(25-6n=1\). This gives \(6n=24\), so \(n=4\). Therefore, the 4th term is 1. The 3rd term is \(25-18=7\), so it is not correct. Exam tip: To find the position of a specified term, equate the nth-term expression to that value and solve for \(n\).

The second differences are (4), and (2n^2+4n+3) gives all starting terms. Half of the second difference gives the coefficient of (n^2).

Given \(a_n=4n+9\), substitute the complete index \(5p-1\) for \(n\): \(a_{5p-1}=4(5p-1)+9=20p-4+9=20p+5\). Therefore, \(20p+5\) is correct. The option \(20p+9\) would result from incorrectly ignoring the \(-1\) in the index. Exam tip: when an index is an expression, substitute the whole expression using brackets.

In (a_n=5n+12), the coefficient of (n) is (5), so the common difference is (5). In a linear (n)th term, the coefficient of (n) gives the difference.

Each term is half of the previous one, so \(a_n=16\left(\frac{1}{2}\right)^{n-1}\). Keep the first term fixed and use exponent (n-1).

The second differences are (6), and (3n^2+4n+4) gives all starting terms. In a quadratic sequence, test options on the first three terms.

The governing concept is constructing an explicit nth-term rule by identifying both the magnitude pattern and the sign pattern. The absolute values of the listed terms are 1, 4, 9, 16, and 25, which are 1², 2², 3², 4², and 5². The signs alternate, beginning with a negative sign: odd-indexed terms are negative and even-indexed terms are positive. Since (−1)ⁿ equals −1 for odd n and +1 for even n, the rule is aₙ = (−1)ⁿn². Checking gives a₁ = −1, a₂ = 4, a₃ = −9, and a₄ = 16, so option C is correct. Option A reverses the sign pattern, option B ignores signs, and option D does not reproduce the sequence.