nth term

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.

This is an arithmetic sequence with first term 16 and common difference 16. The term after 64 is 64 + 16 = 80, so the fifth term is 80. The number 96 is the next, or sixth, term. Exam tip: identify the constant difference between consecutive terms to find the next term.

For the tenth term, substitute \(n=10\): \(a_{10}=30-10=20\). Therefore, the correct answer is \(20\). The value \(21\) would result from incorrectly using \(n=9\). Exam tip: first substitute the given term number carefully into the formula for the \(n\)th term.

The direct answer is option A: \(a_n=4n-1\). This is an arithmetic sequence because the difference between consecutive terms is constant: \(7-3=4\), \(11-7=4\), and \(15-11=4\). The nth-term formula for an arithmetic sequence is \(a_n=a_1+(n-1)d\). Here \(a_1=3\) and \(d=4\), so \(a_n=3+(n-1)4=3+4n-4=4n-1\). Option A is correct. Option B, \(3n+4\), gives 7 for \(n=1\), not 3. Option C, \(4n+3\), gives 7 for the first term, so it starts too high. Option D, \(7n-4\), gives 3 at the first position but gives 10 at the second position, whereas the sequence has 7; it does not preserve the common difference 4. Substitution also confirms option A: for \(n=1,2,3,4\), it gives 3, 7, 11, 15. Memory cue: in an arithmetic sequence, start with the first term and add the common difference \(n-1\) times.

For the third term, substitute n=3. Thus, \(a_3=5^3=5\times5\times5=125\). Therefore, 125 is correct. The value 25 equals \(5^2\), so it is the second term, not the third. Exam tip: substitute the required term number directly for n in the nth-term formula.

The terms form an arithmetic sequence because each term increases by 2. Its first term is a₁ = 17 and its common difference is d = 2. The nth-term formula is aₙ = a₁ + (n − 1)d. For n = 12, a₁₂ = 17 + (12 − 1)2 = 17 + 22 = 39. Therefore option B is correct. Another way to see this is that moving from the first term to the twelfth term requires 11 equal jumps, each of size 2, so the total increase is 22. Option A corresponds to only 10 jumps, while options C and D are too large and would require differences greater than the stated pattern. The answer must be checked against the indexing: the first listed term is the first term, not the zeroth term.

To find the first term, substitute n=1: \(a_1=8(1)+5=13\). Therefore, the correct answer is 13. The value 11 does not result from substituting n=1 in the given formula. Exam tip: to find the first term of a sequence, always put \(n=1\).

The first term is (20) and the difference is (5), so (a_n=5n+15). In exams, check the general term on the first term.

To find the fourth term, substitute \(n=4\): \(a_4=2^4+1=16+1=17\). Therefore, the correct answer is 17. The close distractor 16 results from forgetting to add 1 after evaluating \(2^4\). Exam tip: evaluate the exponent first, then perform the remaining operations.

This is an arithmetic sequence with a negative common difference, because the terms decrease by 10 each time. We have 90−100=−10, 80−90=−10, and 70−80=−10. The nth-term formula for an arithmetic sequence is \\(a_n=a_1+(n-1)d\\). Here, the first term is \\(100\\) and the common difference is \\(−10\\).

Therefore, \\(a_n=100+(n-1)(−10)=100−10n+10=110−10n\\). Hence, option A is correct. Checking the positions makes the result clear: when \\(n=1\\), the expression gives \\(110−10=100\\); when \\(n=2\\), it gives 90; and when \\(n=4\\), it gives 70. The negative sign is essential because the sequence is decreasing. Option B would give 90 at \\(n=1\\), so it cannot represent the sequence.

For the sixth term, substitute \(n=6\): \(a_6=6(6)+6=36+6=42\). Therefore, 42 is correct. The closest distractor, 36, results from missing the final \(+6\). Exam tip: first substitute the term number in the formula, then simplify step by step.

This is an arithmetic progression with first term 2 and common difference 5. Its ninth term is \(a_9=2+(9-1)\times5=42\). The value 37 is the eighth term, so it is a close but incorrect option. In exams, first identify the first term and common difference, then use \(a_n=a+(n-1)d\).

To find the second term, substitute \(n=2\): \(a_2=100+3(2)=100+6=106\). Therefore, 106 is correct. The value 103 is obtained for \(n=1\), so it is the first term, not the second. Exam tip: In nth-term questions, carefully substitute the required term number into the formula.

To find the ninth term, substitute \(n=9\) in the formula: \(a_9=7\times 9-5=63-5=58\). Therefore, 58 is correct. A value such as 56 can result from an error in multiplication or subtraction. Exam tip: first substitute the given value of \(n\), then calculate according to the order of operations.

For the fifth term, substitute n=5. Thus, a_5=5+6=11, so 11 is correct. The value 10 would result from using n=4, which gives the fourth term. Exam tip: In an nth-term question, first substitute the requested term number for n.

For \(a_n=4n+1\), putting \(n=1\) gives the first term as 5, and consecutive terms differ by 4. Hence the sequence is 5, 9, 13, 17, .... Option C has common difference 4 but starts at 1. Exam tip: check \(n=1\) first.

This is an arithmetic progression with first term \(a=4\) and common difference \(d=4\). Therefore, \(a_{12}=a+(12-1)d=4+11\times4=48\). Hence, 48 is correct. The value 44 can result from incorrectly adding only 10 differences; from the first term to the twelfth term, there are 11 differences. Exam tip: use \(a_n=a+(n-1)d\) and substitute the values carefully.

Putting \(n=1\) gives the first term \(4(1)-3=1\), and each next term increases by 4. Hence 1, 5, 9, 13, ... is correct. Option B also has common difference 4, but its first term is 4. In exams, check both the first term and the difference.

For the fifth term, substitute \(n=5\) in the formula: \(a_5=5^2+2=25+2=27\). Hence, 27 is correct. The value 25 is only \(5^2\); the additional 2 must also be included. Exam tip: Substitute the required value of \(n\) first, then follow the order of operations carefully.

The governing concept is identifying the rule of a geometric sequence. Each displayed term is twice the preceding term: 2×2=4, 4×2=8, and 8×2=16. Continuing the same rule gives the fifth term as 16×2=32. Equivalently, the terms can be written as 2¹, 2², 2³, 2⁴, so the fifth term is 2⁵=32. Therefore option B is correct. Option A does not follow the repeated doubling pattern, option C is not produced by multiplying 16 by 2, and option D is the sixth term because 32×2=64. This check also shows why simply adding a fixed number would be inappropriate for this sequence.

Substituting \(n=1,2,3\) in \(2n-1\) gives \(1,3,5\), the odd natural numbers. In contrast, \(2n\) generates even numbers. Exam tip: verify a general term by listing its first three terms.

The governing concept is finding the position of a given value by using the general term. This is an arithmetic sequence with first term 3 and common difference 3, so its nth term is aₙ=3n. To find the position of 36, set 3n=36 and divide both sides by 3: n=12. Hence 36 is the twelfth term, so option C is correct. Direct listing also confirms it: the kth multiple of 3 is 3k, and 36=3×12. Option A gives 3×10=30, option B gives 3×11=33, and option D gives 3×13=39. Those values lie near 36 but do not equal it, so they are plausible counting distractors.