nth term

Each term is halved, so \(a_n=48\left(\frac{1}{2}\right)^{n-1}\). In a geometric sequence, the exponent starts with (n-1).

Given \(a_n=6n^2-11n+9\), \(a_{11}=6(11)^2-11(11)+9=614\) and \(a_5=6(5)^2-11(5)+9=104\). Therefore, \(a_{11}-a_5=614-104=510\). The value 540 results from incorrectly calculating \(a_5\) as 74. Exam tip: substitute each value of \(n\) carefully and find both terms separately before subtracting.

The first term is (27) and the difference is (22), so (a_{20}=27+19\cdot22=445). The (20)th term includes (19) differences.

The terms are (9\cdot1^2,9\cdot2^2,9\cdot3^2,\ldots), so (a_n=9n^2). Identify the coefficient of the square pattern.

Given \(a_n=55-8n\), put \(a_n=7\): \(55-8n=7\). Thus, \(8n=48\), so \(n=6\). Therefore, 7 is the 6th term of the sequence. The 5th term would be 15, not 7. Exam tip: substitute the given term value into the formula for \(a_n\) and solve for \(n\).

The second differences are (12), and (6n^2+9n+1) gives all starting terms. In a quadratic rule, half of the second difference is the coefficient of (n^2).

Given \(a_n=6n+17\), substitute the entire index \(5p-3\) for \(n\): \(a_{5p-3}=6(5p-3)+17=30p-18+17=30p-1\). Hence, \(30p-1\) is correct. \(30p-18\) results if the constant term \(+17\) is not added. Exam tip: always substitute the complete expression in the subscript for \(n\).

The consecutive differences are \(4,8,16,32\), and each difference doubles. Substituting \(n=1,2,3\) in \(2^{n+1}+3\) gives \(7,11,19\), so it matches the sequence. Although \(2^n+5\) gives the first term, it gives \(9\) when \(n=2\), so it is incorrect. Exam tip: for an exponential sequence, first inspect the pattern of consecutive differences.

To obtain the value 160, solve n(n+6)=160. This gives n²+6n−160=0, or (n−10)(n+16)=0. Since n is a positive term number, n=10. The root n=−16 is not valid. Exam tip: for term-number questions, accept only positive integer roots.

In (a_n=15n-23), the coefficient of (n) is (15), so the common difference is (15). In a linear (n)th term, the coefficient of (n) gives the difference.

Each term is multiplied by \(\frac{1}{5}\), so \(a_n=125\left(\frac{1}{5}\right)^{n-1}\). Keep the first term fixed and use exponent (n-1).

The consecutive differences are 7, 9, 11, and 13, increasing by 2 each time. Hence, the nth term should be quadratic. For \(a_n=n^2+4n+3\), we get \(a_1=8\), \(a_2=15\), \(a_3=24\), \(a_4=35\), and \(a_5=48\). Therefore, option A is correct. Option B gives \(a_1=8\) but gives \(a_2=16\), so it fails. Exam tip: Test an nth-term option using the first two or three terms.

Given \(a_n=4n^2-3n+2\), \(a_7=4(7)^2-3(7)+2=177\) and \(a_4=4(4)^2-3(4)+2=54\). Therefore, \(a_7-a_4=177-54=123\). The value 121 may result from an error while multiplying or substituting a term. In exams, evaluate each required term separately before finding their difference.

The successive differences are \(6,9,12,15,\ldots\), so the second differences are constant at \(3\). The terms are three times the triangular numbers: \(3\times1, 3\times3, 3\times6, 3\times10,\ldots\). Since the \(n\)th triangular number is \(\frac{n(n+1)}{2}\), the required term is \(\frac{3n(n+1)}{2}\). For example, \(3n^2\) gives 12 when \(n=2\), whereas the second term is 9. Exam tip: a constant second difference indicates a quadratic sequence.

The numerator (4,7,10,\ldots) gives (3n+1), and the denominator (7,10,13,\ldots) gives (3n+4). Observe numerator and denominator separately.

To obtain the value 149, set \(3n^2+2=149\). Then \(3n^2=147\), so \(n^2=49\) and \(n=7\). Since a term number must be a positive integer, 149 is the 7th term of the sequence. The 8th term is \(3(8)^2+2=194\), so it is not correct. Exam tip: equate the given value to \(a_n\) first, then solve for \(n\).

For a negative term, \(40-5n<0\). This gives \(n>8\), so the smallest integer value is \(n=9\). Checking nearby terms, \(a_8=0\), which is not negative, whereas \(a_9=40-45=-5\). Hence, the 9th term is the first negative term. Exam tip: for the “first” term, choose the smallest positive integer satisfying the inequality.

A quadratic sequence has an \(n^2\) term and constant second differences. For \(a_n=4n^2-3n+1\), the second difference is \(2\times4=8\), which is constant and non-zero. Constant first differences indicate a linear sequence, while a constant ratio is associated with a geometric sequence. Exam tip: for a quadratic term \(an^2\), the constant second difference is \(2a\).

The first differences are \(9,17,25,33\), and their second differences are all \(8\). Hence the sequence has a quadratic nth term, with coefficient of \(n^2\) equal to \(8/2=4\). Substituting \(n=1,2,3\) in \(4n^2-3n+1\) gives \(2,11,28\), respectively, so it is correct. Although \(4n^2-2n\) gives the first term as \(2\), it gives \(12\) as the second term. Exam tip: when second differences are constant, test a form \(an^2+bn+c\) using the initial terms.

Given a_n=2^n+n, substitute n=5: a_5=2^5+5=32+5=37. Therefore, option B is correct. The value 35 could result from adding an incorrect number to 32. Exam tip: When finding an nth term, substitute the value of n in every part of the formula before simplifying the exponent.

Given \(a_n=pn+q\), we have \(a_4+a_7=(4p+q)+(7p+q)=11p+2q=64\) and \(a_5+a_8=13p+2q=76\). Subtracting the first equation from the second gives \(2p=12\), so \(p=6\). Substituting into \(11p+2q=64\) gives \(q=-1\). Hence, \(a_{10}=10(6)-1=59\). The nearby option 61 can result from using an incorrect constant term. Exam tip: subtract paired-term equations first to eliminate \(q\) quickly.

The successive differences are \(10,14,18,22\), and their second differences are constantly \(4\). Hence, the nth term is quadratic. Let \(a_n=2n^2+bn+c\). Substituting \(n=1\) and \(n=2\) gives \(b=4\) and \(c=1\). Therefore, \(a_n=2n^2+4n+1\). Option A gives 7 for \(n=1\), but it gives 16 rather than 17 for \(n=2\). Exam tip: if the constant second difference is \(d\), the coefficient of \(n^2\) is \(d/2\).

Putting \(n=6\) into the formula gives \(a_6=\frac{6(6+5)}{3}=\frac{6\times11}{3}=22\). Therefore, 22 is correct. The close distractor 24 can result from an arithmetic error while adding \(6+5\) or simplifying the division. Exam tip: to find \(a_n\), substitute the value of \(n\) carefully before simplifying multiplication and division.

Here, \(a_{n+1}=3(n+1)^2+5(n+1)=3n^2+11n+8\). Therefore, \(a_{n+1}-a_n=(3n^2+11n+8)-(3n^2+5n)=6n+8\). The option \(6n+6\) results from an incorrect calculation of the constant term. Exam tip: Substitute \(n+1\), expand every bracket, and then subtract \(a_n\).

The first differences are \(5,11,17,23\), and the second differences are all \(6\). Hence the sequence has a quadratic nth term, with coefficient of \(n^2\) equal to \(6/2=3\). For \(a_n=3n^2-4n+1\), we get \(a_1=0\), \(a_2=5\), \(a_3=16\), and \(a_4=33\), so option A is correct. Option C gives \(a_1=0\), but it gives \(a_2=6\), not \(5\). Exam tip: when second differences are constant, start with \(an^2+bn+c\).