nth term

The governing concept is recognizing the square-number sequence. The terms are 1², 2², 3², 4², and so on, so the general rule is a_n = n². Consequently, a_11 = 11² = 121 and a_12 = 12² = 144. Adding the two requested values gives a_11 + a_12 = 121 + 144 = 265. Therefore option C is correct. It is important to calculate the two indexed terms separately; the expression does not mean the 23rd term. The distractors can result from using an incorrect index, squaring only one number, or making an addition error. A useful check is that 121 + 144 is 265, which lies between the neighboring possible values and follows directly from the square-number rule.

Direct answer: aₙ = 3n/2, so option B is correct. Fractions do not change the method: identify how the term depends on its position. The displayed terms are consecutive multiples of 3/2. At position 1, (3/2)×1 = 3/2; at position 2, (3/2)×2 = 3; at position 3, (3/2)×3 = 9/2; and at position 4, (3/2)×4 = 6. Therefore the term at position n is aₙ = (3/2)n = 3n/2. The arithmetic formula gives the same result because a₁ = 3/2 and d = 3/2: aₙ = 3/2 + (n−1)(3/2) = 3n/2. Option A gives 1/3 when n = 1, not 3/2. Option C gives 2 as its first term and has the wrong increase. Option D gives 3 first but then 6, 9, 12, which is twice the required rate. Memory cue: test any proposed formula at n = 1 and n = 2 before accepting it.

The governing concept is identifying an nth-term rule from position-number relationships. The terms are consecutive perfect cubes: 1 = 1³, 8 = 2³, 27 = 3³, and 64 = 4³. Therefore the term in position n is a_n = n³, making option B correct. Substitution verifies the rule: n = 1 gives 1, n = 2 gives 8, n = 3 gives 27, and n = 4 gives 64. Option A gives the square sequence 1, 4, 9, 16. Option C gives 3, 6, 9, 12, while option D gives 2, 4, 8, 16 when n starts at 1. Thus only n³ agrees with every displayed term and with the position pattern.

Direct answer: aₙ = 7n + 1, so option A is correct. The consecutive differences are 15−8 = 7, 22−15 = 7, and 29−22 = 7. Therefore the sequence is arithmetic with first term a₁ = 8 and common difference d = 7. Use aₙ = a₁ + (n−1)d: aₙ = 8 + 7(n−1) = 8 + 7n − 7 = 7n + 1. Substitution confirms the pattern: n = 1 gives 8, n = 2 gives 15, n = 3 gives 22, and n = 4 gives 29. Option B, 8n, starts correctly at 8 but increases by 8, giving 16 next instead of 15. Option C, 7n−1, gives 6 at n = 1. Option D, n+7, gives the first term correctly but increases by only 1. Option A is the only rule satisfying both the initial value and the constant difference. A reliable habit is to check these two conditions rather than checking only one term.

The governing concept is the nth term of an arithmetic sequence. The difference between successive terms is constant: 20−11=9, 29−20=9, and 38−29=9. Therefore a₁=11 and d=9. Applying aₙ=a₁+(n−1)d gives aₙ=11+9(n−1)=11+9n−9=9n+2. Hence option B is correct. Substitution provides a further check: a₁=9(1)+2=11, a₂=20, a₃=29, and a₄=38. Option A has common difference 11. Option C gives 7 as its first term, and option D gives 11 initially but increases by only 2, not by 9. These checks show that only option B fits both the starting value and the growth pattern.

The governing concept is recognizing a quadratic nth-term rule through differences. The first differences are 6, 8, and 10, so they are not constant; however, the second differences are 2 and 2. This suggests a quadratic expression. Test option A: for n=1, 1²+3(1)=4; for n=2, 2²+3(2)=10; for n=3, 3²+3(3)=18; and for n=4, 4²+3(4)=28. Thus aₙ=n²+3n reproduces every given term and option A is correct. Option B and option D are linear, so their first differences would be constant. Option C gives 3, 8, 15, and 24, which does not match the sequence.

The governing concept is evaluating an explicit rule at successive positive integer values of n. The first term is obtained by putting n = 1: a₁ = 2(1) + 5 = 7. The second term is a₂ = 2(2) + 5 = 9, and the third term is a₃ = 2(3) + 5 = 11. Therefore the first three terms are (7, 9, 11), so option B is correct. The constant increase is 2, confirming that these terms form an arithmetic progression. Option A incorrectly treats 5 as the first term, option C uses a constant pattern beginning at 6, and option D begins at 8; none follows the given formula when n starts at 1.

The governing concept is finding the nth term of an arithmetic progression. Here the first term is a₁ = 13 and the common difference is d = 18 − 13 = 5. The nth-term formula is aₙ = a₁ + (n − 1)d. For the eighth term, a₈ = 13 + (8 − 1)5 = 13 + 35 = 48. Therefore option C is correct. A useful check is to continue the sequence: the fifth term is 33, the sixth is 38, the seventh is 43, and the eighth is 48. Option A is actually the seventh term, so it results from using six differences instead of seven. Options B and D do not follow the fixed difference of five and therefore cannot be the eighth term.

The governing concept is the nth-term rule for an arithmetic progression, a_n = a_1 + (n − 1)d. In this sequence, the first term is a_1 = 7 and the common difference is d = 14 − 7 = 7. Therefore, for the tenth term, a_10 = 7 + (10 − 1)7 = 7 + 63 = 70. The same answer is visible because the terms are consecutive multiples of 7: the tenth term is 10 × 7 = 70. Thus option B is correct. Option A, 63, is the ninth term, not the tenth. Option C, 77, is the eleventh term, and option D, 84, is the twelfth term. Correctly counting the position prevents an off-by-one error.

In an arithmetic progression, every term is obtained by adding the same common difference to the preceding term. Here d = 13 − 9 = 4, and the displayed terms are 9, 13, 17, and 21. One more addition of 4 gives the fifth term: 21 + 4 = 25. The formula confirms this result: a_5 = a_1 + (5 − 1)d = 9 + 4 × 4 = 25. Therefore option C is correct. Option A does not preserve the difference of 4. Option B is one less than the required value, while option D is one more; both reflect an arithmetic or continuation error. The fourth term is 21, so exactly one common difference must be added to reach the fifth.

Direct answer: option C, the sixth term. The first term is a₁ = 2 and the common difference is d = 8−2 = 6. To find the position of 32, use aₙ = a₁ + (n−1)d. Substitute: 32 = 2 + (n−1)6. Subtract 2 from both sides: 30 = 6(n−1). Divide by 6: 5 = n−1, so n = 6. A direct list confirms the result: 2, 8, 14, 20, 26, 32. Therefore 32 is the sixth term. Option A is the fourth term, 20; option B is the fifth term, 26; and option D would be the seventh term, 38. The common mistake is to divide 32 directly by 6, forgetting the starting value 2. Always subtract the first term before dividing by the common difference, then add 1 to obtain the position.

The governing concept is the nth-term formula for an arithmetic progression: a_n = a + (n − 1)d. The first term is a = 21, and the common difference is d = 27 − 21 = 6. For the eighth term, substitute n = 8: a_8 = 21 + (8 − 1) × 6 = 21 + 42 = 63. Therefore, option C is correct. The factor 7 appears because there are seven equal jumps from the first term to the eighth term. Option A stops after six jumps, while 61 and 65 do not follow the constant difference of 6. Checking by continuation gives 45, 51, 57, and then 63, which independently confirms the answer.

The governing concept is the constant common difference of an arithmetic progression. Subtract consecutive terms: 19 − 13 = 6, 25 − 19 = 6, and 31 − 25 = 6. Thus every new term is obtained by adding 6 to the preceding term. The fifth term is therefore 31 + 6 = 37, so option C is correct. The nth-term formula confirms the result: a_n = a + (n − 1)d, so a_5 = 13 + (5 − 1) × 6 = 13 + 24 = 37. Option A would increase by only 4 from the fourth term, option B by 5, and option D by 8; none follows the established common difference of 6. Both direct continuation and the formula give the same unambiguous answer.

The governing concept is the nth term of an arithmetic progression, given by a_n = a + (n − 1)d. The first term is a = 18, and the common difference is d = 25 − 18 = 7; the later differences also equal 7. For the eighth term, n = 8, so a_8 = 18 + (8 − 1) × 7 = 18 + 49 = 67. Therefore option B is correct. Continuing the sequence gives 46, 53, 60, and then 67, which independently verifies the calculation. Option A results from using one fewer difference than required, while options C and D add 51 or 53 to the first term instead of the required 49. The first term to the eighth term contains seven equal intervals, not eight, which explains the factor n − 1.

Direct answer: option A, aₙ = 9n − 1. For an AP, aₙ = a + (n − 1)d. The first term is a = 8 and the common difference is d = 17 − 8 = 9. Substitution gives aₙ = 8 + (n − 1)9 = 8 + 9n − 9 = 9n − 1. Check the rule: n = 1 gives 8, n = 2 gives 17, and n = 3 gives 26, exactly the listed terms. B gives 17 when n = 1, so it fails immediately. C also gives 17 at n = 1 and has the wrong constant term. D has a difference of 1 between successive values, not 9. A useful memory cue is first term plus (number of jumps × common difference): there are n − 1 jumps before the nth term.

The governing concept is the general-term formula for an arithmetic progression: a_n = a + (n - 1)d. The first term is a = 11, and the common difference is d = 20 - 11 = 9; the later differences, 29 - 20 and 38 - 29, also equal 9. Substitution gives a_n = 11 + (n - 1)9 = 11 + 9n - 9 = 9n + 2. Therefore option A is correct. Checking n = 1 gives 9(1) + 2 = 11, and checking n = 2 gives 20, so the rule begins exactly as the given sequence. Option C incorrectly adds 11 without adjusting for the n - 1 form, while B and D use coefficients or constants that do not reproduce the terms.

The governing concept is the nth-term rule for a geometric progression: aₙ = arⁿ⁻¹, where a is the first term and r is the common ratio. For the third term, n = 3, so the exponent is 3 − 1 = 2. Hence a₃ = 4 × 5² = 4 × 25 = 100. The same result appears by generating the terms: the first term is 4, the second is 4 × 5 = 20, and the third is 20 × 5 = 100. Therefore, option C is correct. Option A does not result from two multiplications by 5, option B comes from an incorrect calculation, and option D treats the expression as 5³ instead of 4 × 5².

For the fifth term, substitute \(n=5\). Then \(a_5=2(5)+3=10+3=13\). Therefore, 13 is the correct answer. A value such as 15 may result from an error in multiplication or addition. Exam tip: in an \(n\)th-term question, first substitute the given term number carefully for \(n\).

For the fourth term, substitute n=4. Then a_4=5×4−1=20−1=19, so 19 is correct. The value 21 may result from adding 1 instead of subtracting it. Exam tip: after substituting n, perform multiplication before addition or subtraction.

The direct answer is option A: \(a_n=3n\). The sequence is \(3,6,9,12,\ldots\), so each term is a multiple of 3. The first term is \(3=3\times1\), the second is \(6=3\times2\), the third is \(9=3\times3\), and the fourth is \(12=3\times4\). Therefore the term in position \(n\) is \(3\times n\), or \(a_n=3n\). Option A gives exactly this rule. Option B, \(n+3\), gives 4 when \(n=1\), not the first term 3. Option C, \(2n+1\), gives 3 for the first term but gives 5 for the second instead of 6, so it fails. Option D, \(3n-1\), gives 2 for the first term and therefore does not match the sequence. A useful additional check is that the common difference is 3, and an arithmetic sequence beginning at 3 with difference 3 has formula \(a_n=a_1+(n-1)d=3+3(n-1)=3n\). Memory cue: for the positive multiples of 3 beginning with 3, use three times the position number.

Direct answer: aₙ = 3n + 1, so B is correct. Compare consecutive terms: 7 − 4 = 3, 10 − 7 = 3, and 13 − 10 = 3. The difference is constant, so this is an arithmetic sequence. For an arithmetic sequence, aₙ = a₁ + (n − 1)d, where a₁ is the first term and d is the common difference. Here a₁ = 4 and d = 3. Therefore aₙ = 4 + (n − 1)3 = 4 + 3n − 3 = 3n + 1. Check the rule: n = 1 gives 4, n = 2 gives 7, and n = 4 gives 13. A gives 2 when n = 1, so it fails immediately. C gives 4 at n = 1 but then increases by 4, not by 3. D begins with 4 but increases by only 1. Common warning: do not use 3n alone without adjusting for the first term; always test n = 1.

For the sixth term, substitute \(n=6\): \(a_6=10-6=4\). Therefore, the correct answer is 4. Option 5 would result from using \(n=5\), which gives the fifth term. Exam tip: In an nth-term formula, first substitute the correct term number.

For the seventh term, substitute n=7. Thus, a_7=7^2=49, so the correct answer is 49. The value 64 equals 8^2, so it would be the eighth term. Exam tip: substitute the given term number into the nth-term formula before calculating.

The direct answer is option A: \(a_n=3n-1\). The sequence 2, 5, 8, 11 increases by 3 each time, so it is an arithmetic progression with first term \(a_1=2\) and common difference \(d=3\). The nth-term formula for an arithmetic progression is \(a_n=a_1+(n-1)d\). Substituting gives \(a_n=2+(n-1)3=2+3n-3=3n-1\). Checking is useful: for \(n=1\), the formula gives 2; for \(n=2\), it gives 5; for \(n=3\), it gives 8. Option A is therefore correct. Option B, \(3n+2\), gives 5 when \(n=1\), so it starts with the wrong first term. Option C, \(2n+3\), gives 5, 7, 9 for the first three values and does not have the required difference of 3. Option D, \(5n-3\), gives 2 initially but then gives 7, not 5, so its common difference is 5. The common mistake is to identify the difference correctly but forget the \(n-1\) in the formula. Memory cue: first term plus one less than the position multiplied by the common difference.

For the third term, substitute n=3: \(a_3=4\times3+2=12+2=14\). Therefore, 14 is correct. The value 12 is only \(4\times3\); the +2 must also be added. Exam tip: In an nth-term question, first substitute the required value of n correctly into the formula.