nth term

For the fifth term, substitute \(n=5\) in the formula: \(a_5=6\times5-2=30-2=28\). Therefore, 28 is correct. The close distractor 30 results from forgetting to subtract 2. Exam tip: substitute the value of \(n\) first, then follow the correct order of operations.

This is a sequence of powers of (2), so (a_n=2^n). In exams, identify power patterns separately.

To find the fourth term, substitute n=4: \(a_4=20-2(4)=20-8=12\). Therefore, the correct answer is 12. The value 14 is obtained for n=3, so it is the third term. Exam tip: carefully substitute the required term number for n in the nth-term formula.

This is an arithmetic progression with first term 7 and common difference 2. Therefore, the eleventh term is \(a_{11}=7+(11-1)\times2=27\). Hence, 27 is the correct option. In exams, remember to use \(n-1\) in the formula \(a_n=a+(n-1)d\).

For the fourth term, substitute n=4: \(a_4=3^4=3\times3\times3\times3=81\). Therefore, the correct answer is 81. The value 27 equals \(3^3\), so it is the third term. Exam tip: write the required term number in place of n before evaluating the expression.

The sequence consists of successive multiples of 8: 8 × 1, 8 × 2, 8 × 3, and 8 × 4. Thus the term in position n is 8 × n, so the general rule is aₙ = 8n. Option B is therefore correct. The rule can be checked directly: at n = 1 it gives 8, at n = 2 it gives 16, at n = 3 it gives 24, and at n = 4 it gives 32. Option A gives only half of each displayed term, option C does not preserve the constant difference of 8, and option D gives 16 as the first term and doubles the required values. A valid nth-term expression must reproduce every listed position consistently.

In \(a_n=7-3n\), the coefficient of \(n\) is \(-3\). Thus, when \(n\) increases by 1, the term decreases by 3; the common difference is \(-3\). Option A describes an increase. Exam tip: for \(pn+q\), the common difference is \(p\).

This is an arithmetic sequence because the difference between consecutive terms is constant: 15 − 12 = 3, 18 − 15 = 3, and 21 − 18 = 3. For an arithmetic sequence, the nth-term rule is aₙ = a₁ + (n − 1)d. Here a₁ = 12 and d = 3, so aₙ = 12 + (n − 1)3 = 12 + 3n − 3 = 3n + 9. Therefore option A is correct. Substituting n = 1 gives 12, n = 2 gives 15, and n = 3 gives 18. Option B has the wrong growth and first term; option C gives 15 at n = 1; option D gives 12 at n = 1 but increases by 15, not 3.

For the fifth term, substitute n=5: (a_5=5^2+1=25+1=26). Therefore, 26 is the correct answer. 25 is only the square of 5; the +1 from the formula must still be added. Exam tip: To find an nth term, first substitute the term number for n and then simplify.

This is a sequence of multiples of (11), so (a_7=11(7)=77). In exams, identify multiples to solve quickly.

For the third term, substitute \(n=3\): \(a_3=9\times 3-4=27-4=23\). Therefore, the correct answer is 23. The value 21 may result from an error in subtraction. Exam tip: first substitute the term number into the \(n\)th-term formula, then simplify step by step.

This is an arithmetic sequence because every term increases by the same amount. The common difference is 4: 17−13=4, 21−17=4, and 25−21=4. For an arithmetic sequence, the nth term is found by starting with the first term and adding the common difference \\(n-1\\) times. Thus, \\(a_n=a_1+(n-1)d\\), where \\(a_1=13\\) and \\(d=4\\).

Substitution gives \\(a_n=13+(n-1)4=13+4n-4=4n+9\\). Therefore, option A is correct. A quick check confirms it: for \\(n=1\\), the formula gives \\(4(1)+9=13\\); for \\(n=2\\), it gives 17; and for \\(n=4\\), it gives 25. The other expressions do not produce the given first terms in the required order.

For the eighth term, substitute \(n=8\). Thus, \(a_8=\frac{8}{2}=4\). Therefore, the correct answer is 4. Option 2 would result from using \(n=4\), which represents the fourth term. Exam tip: substitute the required term number carefully in an nth-term formula.

For the second term, substitute n=2: \(a_2=15-3(2)=15-6=9\). Therefore, the correct answer is 9. The value 12 is obtained when \(n=1\), so it is the first term, not the second. Exam tip: In an nth-term formula, substitute the required term number carefully for \(n\).

This starts from the second multiple of (7), and the sixth term is (49). In exams, you can also check by counting terms.

To find the first term, substitute \(n=1\) in the formula: \(a_1=4(1)-3=1\). Therefore, 1 is the correct option. Option 3 does not result from substituting \(n=1\); it may be chosen by confusing it with the constant part of the expression. Exam tip: For the first term of any sequence, put \(n=1\).

At (n=1), the term should be (0), so (a_n=2n-2). In exams, be careful with sequences starting from zero.

To find the second term, substitute \(n=2\) in the formula: \(a_2=12\times2+1=24+1=25\). Therefore, the correct answer is 25. The value 24 is only \(12\times2\); the \(+1\) in the formula has not yet been added. Exam tip: For an \(n\)th-term question, first substitute the required value of \(n\) correctly.

The sequence increases by a fixed amount: 18 − 15 = 3, 21 − 18 = 3, and 24 − 21 = 3. Thus it is an arithmetic sequence with first term a₁ = 15 and common difference d = 3. Applying aₙ = a₁ + (n − 1)d gives aₙ = 15 + (n − 1)3 = 15 + 3n − 3 = 3n + 12. Hence option A is correct. A reliable check is to substitute n = 1, which gives 15, and n = 4, which gives 24. Option B gives 15 at n = 1 but then jumps by 15; option C gives 18 as its first term; option D gives the correct second term by coincidence but not the correct constant difference.

For the sixth term, put n=6: a_6=6^2-1=36-1=35. Therefore, 35 is the correct answer. Although 36 is a close option, it is the value before subtracting 1. Exam tip: To find an nth term, substitute the value of n first and then perform the operations in order.