nth term

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 \(an^2+bn+c\).

To find the term number for 131, put a_n=131. Then 7n-9=131, so 7n=140 and n=20. Therefore, 131 is the 20th term. The 19th term is 7(19)-9=124, so it is not correct. Exam tip: To find the position of a given term, substitute its value for a_n and solve for n.

The term can be written as
a_n=n^2-6n+13=(n-3)^2+4
. Since the minimum value of
(n-3)^2
is 0, attained at n=3, the smallest term value is 4. Option 3 is a common confusion because it is the value of n, not the value of the term. Exam tip: rewrite a quadratic in completed-square form to find its minimum quickly.

The numerators \(1,4,9,16\) are \(1^2,2^2,3^2,4^2\), so the numerator of the nth term is \(n^2\). The denominators \(3,8,15,24\) are respectively \(1(1+2),2(2+2),3(3+2),4(4+2)\). Hence, the denominator is \(n(n+2)\). Therefore, the nth term is \(\frac{n^2}{n(n+2)}=\frac{n}{n+2}\). Option B has the numerator pattern but not the required denominator. Exam tip: in fractional sequences, find the numerator and denominator patterns separately before simplifying.

Substituting \(n=4\) into the formula gives \(a_4=5\cdot3^{4-1}-2=5\cdot3^3-2=5\cdot27-2=133\). Therefore, 133 is correct. A value such as 131 can result from an error in evaluating \(3^3\) or in the subsequent multiplication. Exam tip: in expressions with exponents, simplify the exponent first, then evaluate the power and multiplication.

The first differences are \(3,7,11,15\), and the second differences are all \(4\). Hence, the nth term has the quadratic form \(an^2+bn+c\). Since \(2a=4\), we get \(a=2\). Using the first two terms gives \(b=-3\) and \(c=5\), so the nth term is \(2n^2-3n+5\). Option B has the same leading coefficient, but although it gives 4 for \(n=1\), it gives 8 for \(n=2\), not 7. Exam tip: when second differences are constant, assume a quadratic expression and determine its coefficients from initial terms.

Because \(a_n=rn+s\) is linear, the change in the term is proportional to the change in the index. Between the third and tenth terms, the value increases by \(49-14=35\) over \(10-3=7\) index intervals. Thus \(7r=35\), so \(r=5\). Using \(a_3=14\), we obtain \(3(5)+s=14\), hence \(s=-1\). Therefore \(a_{15}=5(15)-1=75-1=74\), so option B is correct. Equivalently, from the tenth to the fifteenth term there are five steps, giving \(49+5\cdot5=74\).

We need \(4n^2-1>100\). Thus, \(4n^2>101\), so \(n^2>25.25\). At \(n=5\), \(a_5=99\), which is not greater than 100, whereas \(a_6=143\). Therefore, the first term greater than 100 is the 6th term. Exam tip: after solving an inequality, check the preceding integer term to confirm that the required term is the first one.

Evaluate the explicit rule separately at the two required indices. For \(n=6\), \((-1)^6=1\), so \(a_6=1(6+2)=8\). For \(n=7\), \((-1)^7=-1\), so \(a_7=-1(7+2)=-9\). Therefore \(a_6+a_7=8+(-9)=-1\), and option A is correct. The key concept is the parity-dependent sign: even indices give a positive factor and odd indices give a negative factor. Options C and D incorrectly combine the magnitudes 8 and 9, while option B reverses the final sign.

Here, \(a_{n-1}=(n-1)^3\). Therefore, \(a_n-a_{n-1}=n^3-(n-1)^3=n^3-(n^3-3n^2+3n-1)=3n^2-3n+1\). Hence, option D is correct. Option A is the result of \((n+1)^3-n^3\), not of the given difference. Exam tip: for the previous term, always replace \(n\) with \(n-1\).

Given \(a_n=2n^2+kn+3\), put \(n=3\): \(a_3=2(3)^2+3k+3=21+3k\). Since \(a_3=30\), \(21+3k=30\), so \(k=3\). Now put \(n=5\): \(a_5=2(5)^2+3(5)+3=50+15+3=68\). Hence, 68 is correct. A value such as 72 may result from using an incorrect value of \(k\) or making an addition error. Exam tip: first use the given term to find the unknown constant, then substitute the required value of \(n\).

The consecutive differences are \(14,22,30,38\), and their common second difference is \(8\). Hence the term has the quadratic form \(an^2+bn+c\), where \(2a=8\), so \(a=4\). Using the first two terms gives \(4n^2+2n\). It gives \(6\) for \(n=1\) and \(20\) for \(n=2\). Option B has the correct leading coefficient but incorrect linear and constant terms. Exam tip: for a constant second difference, the coefficient of \(n^2\) is half the second difference.

In an arithmetic progression, the difference between consecutive terms remains constant: \(a_{n+1}-a_n=d\). Hence, every new term is formed by adding the same fixed number \(d\) to the preceding term. A geometric progression has a constant ratio rather than a constant difference. Exam tip: subtract consecutive terms; if the difference is constant, it is an AP.

The numerator (5,9,13,\ldots) gives (4n+1), and the denominator (6,11,16,\ldots) gives (5n+1). Make separate rules for both sequences.

Given \(a_n=n^2+2n+5\), \(a_9=9^2+2(9)+5=81+18+5=104\), while \(a_1=1^2+2(1)+5=8\). Hence, \(a_9-a_1=104-8=96\). The option 98 can result from an error while evaluating \(a_1\). Exam tip: substitute the value of \(n\) separately in each term before taking their difference.

Interpret the given linear term as \(a_n=an+b\). Then \(a_2+a_9=11a+2b=54\) and \(a_4+a_{11}=15a+2b=78\). Subtracting the first equation from the second gives \(4a=24\), so \(a=6\). Using \(11(6)+2b=54\), we get \(b=-6\). Hence, \(a_{12}=12(6)-6=66\). The nearby option \(63\) does not follow from this linear expression. Exam tip: subtract such pair-sum equations first to eliminate the constant term \(b\).

The first differences are \(11,15,19,23\), and the second differences are all \(4\). Hence the sequence has a quadratic nth term, with coefficient of \(n^2\) equal to \(4/2=2\). Let \(a_n=2n^2+bn+c\). Using \(n=1\) and \(n=2\) gives \(b=5\) and \(c=3\), so \(a_n=2n^2+5n+3\). Option B has the same quadratic coefficient but does not fit the second term. Exam tip: for a constant second difference, use a quadratic form for the nth term.

Given \(a_n=100-7n\), a term smaller than 30 must satisfy \(100-7n<30\). This gives \(-7n<-70\), so \(n>10\). The smallest natural-number value of \(n\) is 11; therefore, the 11th term is the first term less than 30. The 10th term is \(a_{10}=30\), which is not less than 30. Exam tip: for “less than,” use \(<\); an equal value is not included.