nth term

The key concept is recognizing a sequence whose terms are consecutive multiples of 5. The first term is 5, the second is 5 × 2, the third is 5 × 3, and the fourth is 5 × 4. Therefore the term in position n is 5 × n, or aₙ = 5n. Substitution verifies the rule: n = 1 gives 5, n = 2 gives 10, n = 3 gives 15, and n = 4 gives 20. Hence option B is correct. Option A gives 6 as the first term, option C gives 10 as the first term and grows twice as fast, and option D produces square numbers rather than the displayed multiples of 5. The rule must work for every position, not just one term.

For the eighth term, substitute \(n=8\): \(a_8=2(8)-1=16-1=15\). Therefore, the correct answer is \(15\). \(16\) would result if the \(-1\) in the formula were omitted. Exam tip: in an \(n\)th-term question, substitute the given value of \(n\) carefully into the formula.

Direct answer: the tenth term is 28, so C is correct. The sequence increases by 3 each time: 4 − 1 = 3, 7 − 4 = 3, and 10 − 7 = 3. It is therefore an arithmetic sequence with first term a₁ = 1 and common difference d = 3. The nth-term formula is aₙ = a₁ + (n − 1)d. For n = 10, a₁₀ = 1 + (10 − 1)3 = 1 + 27 = 28. Equivalently, the rule simplifies to aₙ = 3n − 2, and 3(10) − 2 = 28. Option A, 25, is the ninth term because 1 + 8×3 = 25. B does not fit the repeated difference of 3. D, 30, comes from multiplying the position by 3 and forgetting that the sequence starts at 1 rather than 3. Memory cue: from the first term to the tenth there are nine jumps, not ten jumps.

For the sixth term, substitute \(n=6\) in the formula: \(a_6=7\times6=42\). Therefore, 42 is the correct answer. The value 49 would be obtained for \(n=7\), so it is the seventh term. Exam tip: In an nth-term question, carefully substitute the required term number into the formula.

The sequence is arithmetic because each term increases by the constant difference 5. With first term a₁ = 6 and common difference d = 5, the nth-term formula is aₙ = a₁ + (n − 1)d. Substitution gives aₙ = 6 + 5(n − 1) = 6 + 5n − 5 = 5n + 1. Therefore option A is correct. Verification is important: n = 1 gives 5(1) + 1 = 6, n = 2 gives 11, and n = 4 gives 21. Option B gives 5 when n = 1, option C gives 4, and option D has a common difference of only 1. These checks show why the selected expression is the only one matching all displayed terms.

For the seventh term, substitute \(n=7\): \(a_7=3\times 7+4=21+4=25\). Therefore, the correct answer is \(25\). \(23\) may result from an error in multiplication or addition. Exam tip: after substituting the term number, multiply first and then add.

This is a sequence of multiples of (10), so (a_n=10n). In exams, identify a sequence of multiples quickly.

For the ninth term, substitute \(n=9\) in the rule: \(a_9=9+8=17\). Therefore, the correct answer is 17. The value 16 would result from using \(n=8\), which gives the eighth term. Exam tip: always substitute the term number asked for in place of \(n\).

The difference is (5), and the rule must give (9) at (n=1), so (a_n=5n+4). In exams, check the rule on the first two terms.

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.

Direct answer: Option B, \\(a_n=2^n\\), is correct. This is a geometric sequence: each term is obtained by multiplying the previous term by 2. Check the positions carefully: first term is 2, and \\(2^1=2\\); second is 4, and \\(2^2=4\\); third is 8, and \\(2^3=8\\); fourth is 16, and \\(2^4=16\\). Thus the nth term is \\(2^n\\). Option A, \\(2n\\), gives 2, 4, 6, 8, so it stops matching at the third term. Option B matches every displayed term and is correct. Option C, \\(n^2\\), gives 1, 4, 9, 16, so it also fails. Option D, \\(2^{n-1}\\), gives first term 1 instead of 2; it would be correct for a sequence beginning with 1. Memory cue: match the first term with the exponent’s starting value.

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 \(p\).

Direct answer: aₙ = 3n + 9, so A is correct. Subtract neighboring terms: 15 − 12 = 3, 18 − 15 = 3, and 21 − 18 = 3. The constant difference shows that the sequence is arithmetic. Use aₙ = a₁ + (n − 1)d. With a₁ = 12 and d = 3, aₙ = 12 + (n − 1)3 = 12 + 3n − 3 = 3n + 9. Testing confirms it: at n = 1, the rule gives 12; at n = 2, it gives 15; and at n = 3, it gives 18. B gives 12 at n = 1 but then grows by 12, so it cannot match. C gives 15 for the first term, not 12. D gives a first term of 12 but has common difference 15, not 3. A useful check is to verify both the first term and the difference; matching only one term is not enough.

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 \(n=1\).