nth term

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 difference is (4) and the first term is (3), so (a_n=4n-1). In exams, check the first term before choosing an option.

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.

The consecutive differences are 9 − 4 = 5, 14 − 9 = 5, and 19 − 14 = 5, so the sequence is arithmetic with first term a₁ = 4 and common difference d = 5. The governing formula is aₙ = a₁ + (n − 1)d. Substituting the values gives aₙ = 4 + (n − 1)5 = 4 + 5n − 5 = 5n − 1. Therefore option A is correct. Verification is important: n = 1 gives 4, n = 2 gives 9, and n = 4 gives 19. Option B gives 6 as its first term; option C begins with 9 and has the wrong difference; option D has a difference of only 1. Thus only option A reproduces both the starting value and the constant increase.

To find the eighth term, substitute \(n=8\) in \(a_n=3n+2\): \(a_8=3\times8+2=24+2=26\). Hence, \(26\) is correct. \(24\) is only \(3\times8\); the constant term \(+2\) must also be added. Exam tip: After substituting the term number for \(n\), include every constant term and sign carefully.

Here \(a=18\) and \(d=-4\). Using \(a_n=a+(n-1)d\) gives \(18-4(n-1)\). The expression \(18-4n\) gives 14 when \(n=1\), not the first term 18. Exam tip: always test a proposed nth-term rule with \(n=1\).

In this geometric progression, the first term is \(a=2\) and the common ratio is \(r=6/2=3\). The \(n\)th term is \(a_n=ar^{n-1}\). Therefore, \(a_6=2\times 3^{6-1}=2\times 3^5=486\). Although 243 equals \(3^5\), it misses multiplication by the first term, 2. Exam tip: In a GP, the exponent is \(n-1\), not \(n\).

The displayed terms are 1, 4, 9, and 16. These are the squares of the counting numbers: \\(1=1^2\\), \\(4=2^2\\), \\(9=3^2\\), and \\(16=4^2\\). This pattern continues, so the term in position \\(n\\) is the square of \\(n\\). In general, this gives \\(a_n=n^2\\).

For example, putting \\(n=1\\) gives 1, putting \\(n=2\\) gives 4, and putting \\(n=4\\) gives 16, exactly matching the sequence. Option A, \\(n^2\\), is therefore correct. The expression \\(2n\\) would describe an even-number pattern, while \\(3n-2\\) gives an arithmetic sequence, so neither reproduces all the given square numbers.

Substitute n=5 in the rule: a_5=2(5)^2+1. First, 5^2=25; then 2×25=50, and 50+1=51. Hence, the correct answer is 51. A value such as 41 can result from an error in squaring or multiplication. Exam tip: substitute the value of n into the complete formula and evaluate powers first.

This is an arithmetic progression with first term 6 and common difference 5. Therefore, its nth term is \(a_n=6+(n-1)\times5=5n+1\). Substituting \(a_n=46\) gives \(5n+1=46\), so \(5n=45\) and \(n=9\). Hence, 9 is the correct option. The 8th term is 41, not 46. Exam tip: first identify the common difference, then use \(a_n=a+(n-1)d\).