nth term

The first term is (15) and the difference is (-4), so (a_n=15+(n-1)(-4)=19-4n). Pay attention to the sign in a decreasing sequence.

Given \(a_n=n(n+1)\). Substituting \(n=6\), we get \(a_6=6(6+1)=6\times7=42\). Therefore, 42 is the correct option. 48 would result from using \(6\times8\), which does not follow the given formula. Exam tip: Substitute the value of \(n\) carefully before simplifying the expression.

The common ratio is (5), so (a_4=4\cdot5^3=500). In a geometric term, the exponent of the ratio is (n-1).

Given \(a_n=5n+6\), \(a_7=5\times7+6=41\) and \(a_3=5\times3+6=21\). Therefore, \(a_7-a_3=41-21=20\). The value \(25\) would result if the difference between the term numbers were 5, but here \(7-3=4\). Exam tip: for a linear sequence \(a_n=dn+c\), use \(a_p-a_q=d(p-q)\) to find the difference quickly.

Compare the terms in order: \(4=1^2+3\), \(7=2^2+3\), \(12=3^2+3\), and \(19=4^2+3\). Therefore, the \(n\)th term is \(a_n=n^2+3\). The option \(n^2+2\) is close, but it gives \(3\) as the first term instead of \(4\). Exam tip: for such sequences, compare the first few terms with \(1^2,2^2,3^2,\ldots\) and check for a constant difference.

For an AP, \(a_n=a+(n-1)d\). Here \(a=20\) and \(d=-3\), so \(a_n=20-3(n-1)\). Formula \(20-3n\) gives 17 as the first term. Exam tip: substitute \(n=1\) to check a proposed nth-term formula.

The terms are (9,18,36,72,144,288), so (288) is the sixth term. For small questions, listing terms is quick.

Given \(a_n=\frac{n(n+1)}{2}\). Substituting \(n=8\), we get \(a_8=\frac{8(8+1)}{2}=\frac{8\times9}{2}=36\). Therefore, option C is correct. A value such as \(40\) can result from substituting incorrectly in the formula. Exam tip: replace \(n\) with the given term number before calculating.

This is an arithmetic progression with first term \(a=20\) and common difference \(d=17-20=-3\). Its \(n\)th term is \(a_n=a+(n-1)d\). Hence, \(a_8=20+(8-1)(-3)=20-21=-1\). Therefore, -1 is correct. The value 0 can result from incorrectly using the term number or forgetting the \((n-1)\) factor. Exam tip: In \(a_n\), add the common difference \((n-1)\) times, not \(n\) times.

Given \(a_n=3n^2-2n\), substitute \(n=4\): \(a_4=3(4)^2-2(4)=3\times16-8=48-8=40\). Therefore, 40 is correct. The value 36 can result from an error while evaluating \(2\times4\) or during subtraction. Exam tip: substitute the term number first, then follow the order of powers, multiplication, and subtraction.

The governing concept is an explicit or general rule for a sequence. Rewrite the given terms as products: 2 = 1×2, 6 = 2×3, 12 = 3×4, and 20 = 4×5. The kth term is therefore k(k+1), so the nth term is aₙ = n(n+1) = n²+n. Checking confirms the rule: for n=1 it gives 2, for n=2 it gives 6, for n=3 it gives 12, and for n=4 it gives 20. Thus option A is correct. Option B gives 4, 6, 8, ...; option C gives 3, 6, 11, ...; and option D gives 2, 5, 8, ..., so none reproduces the complete sequence.

Given \(a_n=2n+9\) and \(a_n=35\), we get \(2n+9=35\). Subtracting 9 gives \(2n=26\), so \(n=13\). Thus, 35 is the 13th term of the sequence. For the close distractor 12, \(a_{12}=2(12)+9=33\), not 35. Exam tip: To find a term’s position, substitute its value for \(a_n\) and solve the resulting equation.

The governing concept is the nth term of a geometric progression. In a geometric progression, each term is obtained by multiplying the preceding term by a constant common ratio. Here the ratio is 32/64 = 1/2. Starting from the first term and applying this ratio gives a₁=64, a₂=32, a₃=16, a₄=8, a₅=4, and a₆=2. Equivalently, use aₙ = arⁿ⁻¹: a₆ = 64(1/2)⁵ = 64/32 = 2. Therefore option B is correct. Option D is the fourth term, option C is the fifth term, and option A would be the seventh term, so these distractors result from stopping one or more steps too early or continuing one step too far.

This is an arithmetic progression with first term \(a=7\) and common difference \(d=7\). Therefore, \(a_n=a+(n-1)d=7+(n-1)\times7=7n\). Substituting \(n=1\) gives \(7\), and substituting \(n=4\) gives \(28\), so \(7n\) is correct. The expression \(n+7\) increases by only 1 each time, so it does not represent this sequence. Exam tip: use \(a_n=a+(n-1)d\) for an arithmetic progression.

To find the fifth term, substitute n=5 in the formula: \(a_5=9-4(5)=9-20=-11\). Therefore, the correct answer is -11. The value -9 would result from incorrectly subtracting 18 from 9 instead of calculating 4×5 correctly. Exam tip: For an nth-term formula, substitute the value of n first, multiply, and then perform addition or subtraction.

The successive differences are 5, 7, and 9, which are consecutive odd numbers, so the pattern is quadratic. Substituting \(n=1,2,3,4\) into \(n^2+2n\) gives 3, 8, 15, and 24 respectively. Therefore, option A is correct. Option B gives 6 when \(n=2\), not 8. Exam tip: verify an nth-term formula by checking it against the first few terms.

Substituting \(n=4\), \(a_4=2\cdot5^{4-1}=2\cdot5^3=2\cdot125=250\). Therefore, the correct answer is 250. The exponent is \(n-1\), so it must be evaluated as \(4-1=3\); using \(5^4\) would give an incorrect value. Exam tip: substitute the term number into the exponent carefully before calculating.

This is an arithmetic sequence with first term 11 and common difference 4. Therefore, \(a_n=11+(n-1)\times4=4n+7\). Substituting \(a_n=63\) gives \(4n+7=63\), so \(4n=56\) and \(n=14\). Hence, 14 is the correct option. The 15th term is 67, so 15 is a close but incorrect distractor. Exam tip: To find a term position, substitute the given term in the formula for \(a_n\) and solve for \(n\).

Given (a_n=n^2-1), substitute n=9 to find the ninth term: (a_9=9^2-1=81-1=80). Therefore, 80 is correct. The value 81 is only the square of 9; subtracting 1 is still required. In exams, substitute the term number first and then follow the order of operations carefully.

In this GP, the first term is \(a=1\) and the common ratio is \(r=4\). Therefore, the \(n\)th term is \(T_n=ar^{n-1}=4^{n-1}\). Since \(1024=4^5\), we get \(n-1=5\), so \(n=6\). Hence, 1024 is the sixth term. The fifth term is \(4^4=256\), not 1024. Exam tip: In a GP, compare the exponent with \(n-1\) to find the term position.