nth term

The consecutive terms satisfy \(a_{n+1}=2a_n+4\): \(4\to12\to28\to60\). The corresponding nth term is \(a_n=2^{n+2}-4\). Checking, for \(n=1\), it gives \(8-4=4\), and for \(n=2\), it gives \(16-4=12\). The option \(2^{n+2}+4\) gives 12 as the first term, so it is incorrect. Exam tip: substitute the first two values of \(n\) to check a proposed sequence formula quickly.

For positive integer values of n, \(a_n=n^2-3n+5=(n-1)(n-2)+3\). The product \((n-1)(n-2)\) is 0 at n = 1 and n = 2, and it is positive for other positive integers. Hence, the smallest term value is \(3\). Option \(2\) may seem tempting because the quadratic has its real minimum at \(n=\tfrac32\), but a sequence index must be an integer. Exam tip: while finding extrema of a sequence, use only valid integer indices.

The required condition is \(9-2n<-15\). This gives \(-2n<-24\), so \(n>12\). Since \(n\) must be a positive integer, the smallest value greater than 12 is 13. Therefore, the 13th term is the first term less than \(-15\). The 12th term equals \(-15\), so it is not correct. Exam tip: for “less than,” retain the strict inequality \(<\).

The successive differences are \(3,5,7,9\), which are consecutive odd numbers. The successive differences of \(n^2\) also follow \(3,5,7,\ldots\), so the term has the form \(n^2+c\). Using the first term, \(1^2+c=5\), giving \(c=4\). Hence, the \(n\)th term is \(n^2+4\). The option \(n^2+3\) gives 4 as the first term, so it is incorrect. Exam tip: for a sequence involving squares, check the first and second differences.

Given \(a_n=rn+s\), we have \(a_{11}-a_6=5r\). Thus, \(57-32=25=5r\), so \(r=5\). Using \(a_6=6(5)+s=32\), we get \(s=2\). Hence, \(a_{20}=20(5)+2=102\). Option 100 ignores the constant term \(s\). Exam tip: subtracting two terms eliminates \(s\) and helps find \(r\) quickly.

Given \(a_n=n^3\), we get \(a_5=5^3=125\) and \(a_3=3^3=27\). Therefore, \(a_5-a_3=125-27=98\). Option 96 may result from an error in evaluating the cubes or in subtraction. Exam tip: find each term separately before taking their difference.

Given \(a_n=4n^2+n\), we have \(a_{n+1}=4(n+1)^2+(n+1)\). Hence, \(a_{n+1}-a_n=4[(n+1)^2-n^2]+1=4(2n+1)+1=8n+5\). Therefore, \(8n+5\) is correct. A result such as \(8n+3\) can arise from an error in expanding the constant terms. Exam tip: while finding \(a_{n+1}\), replace every occurrence of \(n\) with \(n+1\).

The correct answer is option B: \\(a_n=\\frac{n(n+1)(2n+1)}{6}\\). Look at the sequence term by term: 1 is \\(1^2\\); 5 is \\(1^2+2^2=5\\); 14 is \\(1^2+2^2+3^2=14\\); 30 adds \\(4^2\\); and 55 adds \\(5^2\\). Thus the nth term is the sum of the first n squares. The standard formula for this sum is \\(1^2+2^2+\cdots+n^2=\\frac{n(n+1)(2n+1)}{6}\\). Option A is the sum of the first n integers, so it gives 3 for n=2 instead of 5. Option B gives 5 for n=2 and 14 for n=3, so it matches. Option C gives \\(n^3-1\\), which gives 7 for n=2, not 5. Option D gives \\(n^2+n-1\\), which gives 5 at n=2 but 11 at n=3, not 14. Check several terms, not just one. Memory cue: consecutive differences are \\(4,9,16,25\\), the squares.

The given terms are \(3^2,4^2,5^2,6^2,7^2\), respectively. For term number \(n=1\), the base of the square is \(3\), so the base follows the pattern \(n+2\). Hence, the nth term is \((n+2)^2\). For example, at \(n=2\), it gives \((2+2)^2=16\). The expression \(n^2+8\) gives 12 for the second term, so it is not correct. Exam tip: write the initial terms as perfect squares and observe the sequence of their square roots.

Putting n = 5, a_5 = 2(5) + (-1)^5 = 10 - 1 = 9. Since 5 is odd, (-1)^5 = -1. The value 11 would result from incorrectly taking (-1)^5 as +1. Exam tip: remember that (-1)^n is +1 for even n and -1 for odd n.

To find the position of 55, put \(a_n=55\): \(6n-11=55\). Thus, \(6n=66\), so \(n=11\). Therefore, 55 is the 11th term of the sequence. For example, putting \(n=10\) gives \(49\), not 55. Exam tip: To find the term number of a given value, equate \(a_n\) to that value and solve for \(n\).

Given \(a_n=an+b\), we have \(a_2+a_5=(2a+b)+(5a+b)=7a+2b=29\) and \(a_3+a_6=9a+2b=39\). Subtracting the first equation from the second gives \(2a=10\), so \(a=5\). Substituting in \(7a+2b=29\) gives \(b=-3\). Hence, \(a_{10}=10(5)-3=47\). The nearby option 45 is incorrect because it would require \(b=-5\). Exam tip: subtract such paired-sum equations first, as the constant term often cancels out.

The governing concept is recognizing the formula for tetrahedral numbers, which are cumulative sums of triangular numbers. Test option B at successive values of n: for n = 1, 1·2·3⁄6 = 1; for n = 2, 2·3·4⁄6 = 4; for n = 3, 3·4·5⁄6 = 10; for n = 4, 4·5·6⁄6 = 20; and for n = 5, 5·6·7⁄6 = 35. It therefore matches every listed term, so option B is correct. The first differences are 3, 6, 10, and 15, themselves triangular numbers, which supports the same pattern. Option A gives only triangular numbers, while options C and D fail when checked at later positions.

For \(a_n=n^2+1\), \(a_{n+1}-a_n=(n+1)^2+1-(n^2+1)=2n+1\). The difference between consecutive first differences is therefore \(2\), which is constant. \(3n-2\) is linear: its first differences are already constant and its second difference is \(0\), so it is not a quadratic function. Exam tip: constant non-zero second differences usually indicate a quadratic sequence.

The governing concept is using finite differences to identify a quadratic nth-term rule. The first differences are 7, 11, 15, and 19; their second differences are 4, 4, and 4. For a quadratic an² + bn + c, the constant second difference is 2a, so 2a = 4 and a = 2. Testing the remaining coefficients gives option B: at n = 1, 2 + 3 + 1 = 6; at n = 2, 8 + 6 + 1 = 15, which does not match the sequence. Therefore the supplied sequence and option B are inconsistent. In fact, solving from the terms gives aₙ = 2n² + n + 3: it yields 6, 13, 24, 39, 58. Thus option A is the unambiguous correct answer; option B is not correct.

We need \(a_n<10\). Thus, \(50-4n<10\), so \(-4n<-40\) and hence \(n>10\). Since \(n\) must be an integer, the smallest possible value is \(11\). In fact, \(a_{10}=10\), which is not less than 10, whereas \(a_{11}=6\). Exam tip: use \(<\), not \(\leq\), for “less than.”

The numerators \(2,5,8,11,\ldots\) form an arithmetic progression with nth term \(2+(n-1)\times3=3n-1\). The denominators \(5,8,11,14,\ldots\) similarly have nth term \(5+(n-1)\times3=3n+2\). Therefore, the nth term of the given sequence is \(\frac{3n-1}{3n+2}\). For example, option D gives \(\frac{2}{4}\) at \(n=1\), not the first term \(\frac{2}{5}\). Exam tip: in fractional sequences, determine the numerator and denominator patterns separately.

Subtracting the two sums gives (2a=12), so (a=6). Then (12a+2b=74) gives (b=1), so (a_1=a+b=7).

The first term is (17) and the difference is (8), so (a_n=17+(n-1)8=8n+9). In an arithmetic sequence, (n-1) differences are added.

Substitute n=6: a_6=7(6)^2-3(6)+5=7×36-18+5=252-18+5=239. Therefore, 239 is correct. Option 252 is only the value of 7×6² and ignores the terms -3n+5. Exam tip: evaluate the exponent first, then perform multiplication and addition or subtraction.

The successive differences are \(17,25,33,41\), and their second differences are all \(8\). Hence the nth term is quadratic: \(a_n=4n^2+bn+c\). Substituting \(n=1\) and \(n=2\) gives \(b=5\) and \(c=0\). Therefore, \(a_n=4n^2+5n\). Option C gives 9 for \(n=1\), but it gives 25 rather than 26 for \(n=2\). Exam tip: when second differences are constant, test a form \(an^2+bn+c\).

Given \(a_n=2n^2+9n-11\), \(a_{10}=2(10)^2+9(10)-11=279\) and \(a_4=2(4)^2+9(4)-11=57\). Therefore, \(a_{10}-a_4=279-57=222\). The value 204 results from incorrectly calculating \(a_4\) as 75. Exam tip: substitute each value of \(n\) separately before finding the difference.

The second differences are (8), and (4n^2+1) gives all starting terms. Half of the second difference is the coefficient of (n^2).

Given \(a_n=35-6n\), put \(a_n=-43\): \(35-6n=-43\). Thus, \(-6n=-78\), so \(n=13\). Therefore, \(-43\) is the 13th term. The nearby distractor, the 14th term, is incorrect because \(a_{14}=35-84=-49\). Exam tip: take special care with signs when transposing negative terms in an equation.

The direct answer is option B,
\(-42\). Start with the given rule \(a_n=(-1)^n(5n-3)\). Put \(n=9\): first, \(5(9)-3=45-3=42\). Next, because 9 is odd, \((-1)^9=-1\). Therefore \(a_9=(-1)(42)=-42\). Option A, \(42\), has the correct magnitude but misses the negative sign. Option B is correct because it uses both the bracket value and the odd-position sign. Option C, \(48\), comes from an arithmetic error in \(5(9)-3\). Option D, \(-48\), has the right negative sign but the wrong arithmetic value. Always calculate the ordinary expression first and then apply the alternating sign. Memory cue: odd index gives a minus sign here; even index gives a plus sign.