Finding the $n$th term of an AP

Correct answer: B. 54

The row counts form an arithmetic progression because every successive row increases by the same number, 3. The first term is a = 12, the common difference is d = 3, and the required row number is n = 15. Apply aₙ = a + (n − 1)d: a₁₅ = 12 + (15 − 1)×3 = 12 + 14×3 = 12 + 42 = 54. Therefore option B is correct. In this context, the row number corresponds to the term number. Option A uses only thirteen or fewer increases, option C adds one extra increase, and option D contradicts the exact calculation. The sequence begins 12, 15, 18 and reaches 54 at the fifteenth term.

Correct answer: C. 27

The daily numbers form an AP: the first day's value is a = 5 and the fixed daily increase is the common difference d = 2. The day number is the term number, so for the 12th day use aₙ = a + (n − 1)d. Thus a₁₂ = 5 + (12 − 1)×2 = 5 + 11×2 = 5 + 22 = 27. Hence option C is correct. Eleven increases are needed after the first day to reach day twelve. Option A is too small, option B adds one extra increase, and option D does not follow the fixed increment of two. Listing the values as 5, 7, 9, ... also leads to 27 on day twelve.

Correct answer: D. 150 cm

The step heights form an AP because each step increases by the same amount. The first height is a = 15 cm, the common difference is d = 15 cm, and the required term number is n = 10. Using aₙ = a + (n − 1)d, we obtain a₁₀ = 15 + (10 − 1)×15 = 15 + 9×15 = 15 + 135 = 150 cm. Therefore option D is correct. There are nine increases after the first step, not ten. Option A represents only the added part, option B adds one extra increase, and option C comes from an incorrect multiplication or addition. Repeatedly adding 15 gives 15, 30, 45, ..., 150 for the tenth step.

Correct answer: A. First term 12, common difference −3

The standard form is \(a_n=a+(n-1)d\). Comparing it with \(12-3(n-1)\) gives \(a=12\) and \(d=-3\). The negative difference means each next term decreases by 3. Exam tip: carefully retain the sign of the coefficient of \((n-1)\).

Correct answer: B. (-60)

The direct answer is B, -60. In an arithmetic progression, the nth-term formula is \(a_n=a+(n-1)d\), where a is the first term and d is the common difference. Here the first term is 12, and each term decreases by 3, so \(d=9-12=-3\). Thus \(a_{25}=12+(25-1)(-3)=12+24(-3)=12-72=-60\). Option A, 60, has the wrong sign and ignores that the sequence is decreasing. Option B, -60, is correct. Option C, -57, results from using 25 instead of 24 differences; from the first term to the 25th term there are always \(n-1=24\) steps. Option D, -63, uses one extra decrease. The key memory rule is “nth term means n minus 1 jumps from the first term,” and a negative common difference must retain its negative sign.

Correct answer: C. (15)

The direct answer is option C, 15. In an arithmetic progression, consecutive terms change by the same common difference d. The nth-term formula is a_n=a+(n-1)d, where a is the first term. Here a=5, d=7, and a_n=103. Substitute these values: 103=5+(n-1)7. Subtract 5 from both sides: 98=7(n-1). Divide by 7: 14=n-1. Add 1: n=15. Therefore, 103 is the fifteenth term. Option A, 13, would give the term 5+12(7)=89, not 103. Option B, 14, would give 5+13(7)=96, so it is one term too early. Option C, 15, gives 5+14(7)=5+98=103, exactly as required. Option D, 16, would give 5+15(7)=110, which is too large. The common mistake is to divide 98 by 7 and stop at 14; that number is n-1, not n. Always add 1 after finding n-1. A useful check is to list the pattern: the first term is 5, and reaching 103 requires fourteen jumps of 7, so the term number is fourteen jumps plus the first term, namely 15.

Correct answer: D. -2

For an AP, the sixth term is \(a_6=a+5d\). Thus, \(23=a+5(5)=a+25\), so \(a=-2\). If the first term were 2, the sixth term would be \(2+25=27\), so that option is incorrect. Exam tip: in the \(n\)th-term formula, the coefficient of \(d\) is always \(n-1\).

Correct answer: A. 18th

The governing nth-term formula is aₙ = a + (n − 1)d. In the given AP, the first term is a = 3 and the common difference is d = 8 − 3 = 5. Set the nth term equal to the target value: 88 = 3 + (n − 1)5. Thus 85 = 5(n − 1), so n − 1 = 17 and n = 18. Therefore 88 is the 18th term, making option A correct. The use of n − 1 is essential because the first term is obtained after zero differences, not one. A check gives a₁₈ = 3 + 17 × 5 = 88, while a₁₇ = 83 and a₁₉ = 93. Hence the target appears exactly at the stated position and the other choices are excluded.

Correct answer: B. 146

The first term of the AP is \(a=-4\), and its common difference is \(d=1-(-4)=5\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{31}=-4+(31-1)\times5=-4+150=146\). Hence, 146 is the correct answer. Using \(31\times5\) instead of \((31-1)\times5\) would give 151, which is incorrect. Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

Correct answer: C. (4)

The difference of terms is (81-41=40) and the difference of positions is (10), so (d=4). Use (d=\frac{\text{difference of terms}}{\text{difference of positions}}).

Correct answer: C. (78)

(d=\frac{46-14}{12-4}=4), so (a_{20}=46+8\times4=78). Moving from the nearest known term is a faster method.

Correct answer: D. -12

In this AP, the first term is \(a=100\) and the common difference is \(d=92-100=-8\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{15}=100+(15-1)(-8)=100-112=-12\). Hence, \(-12\) is correct. An answer such as \(-4\) can result from using the wrong term count instead of \(n-1\). Exam tip: always check the negative sign of the common difference.

Correct answer: A. 5

For an AP, the nth-term formula is \(a_n=a+(n-1)d\). Here, \(a=18\) and \(a_{16}=93\), so \(93=18+15d\). Thus, \(15d=75\), giving \(d=5\). If the common difference were 4, the 16th term would be \(18+15\times4=78\), not 93. Exam tip: use \(n-1\), not \(n\), in the nth-term formula.

Correct answer: B. 95

For any two terms of an AP, \(a_n=a_m+(n-m)d\). Therefore, \(a_{17}=a_3+(17-3)d=11+14\times6=95\). Hence, 95 is correct. Option 83 uses \(12\times6\), but there are 14 common differences from the 3rd term to the 17th term. Exam tip: always use the difference of the term indices, \(n-m\).

Correct answer: B. (50)

Here (a=2.5), (d=2.5), so (a_{20}=2.5+19\times2.5=50). In decimal APs, keep multiplication exact.

Correct answer: B. 119

Given \(a_n=4n-1\), substitute \(n=30\) to find the 30th term: \(a_{30}=4(30)-1=120-1=119\). Therefore, 119 is correct. The value 121 would result from adding 1 instead of subtracting it, so it is not correct. Exam tip: in a direct \(n\)th-term formula, substitute the required term number carefully for \(n\).

Correct answer: A. \(-59\)

Using the given direct formula \(a_n=7-3n\), substitute \(n=22\): \(a_{22}=7-3(22)=7-66=-59\). Hence, \(-59\) is correct. An answer such as \(-57\) can result from an error in multiplying \(3\times22\) or in subtraction. Exam tip: In a direct formula, substitute the term number for \(n\) and perform multiplication before subtraction.

Correct answer: C. (20)

Here (a=\frac{1}{2}), (d=\frac{1}{2}), so (a_{40}=\frac{1}{2}+39\cdot\frac{1}{2}=20). Fractions can also be checked as decimals.

Correct answer: C. 7

For an AP, \(a_n=a_1+(n-1)d\). Thus, \(35=a_1+(8-1)\times4=a_1+28\), so \(a_1=7\). If 9 were chosen, the eighth term would be \(9+28=37\), not 35. Exam tip: always use \((n-1)d\) in the formula for the \(n\)th term.

Correct answer: C. 17th term

Here, the first term is \(a=15\) and the common difference is \(d=21-15=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(111=15+(n-1)\times6\), so \(96=6(n-1)\) and \(n=17\). Therefore, 111 is the 17th term. The 16th term is \(105\), so it is close but not correct. Exam tip: while finding a term number, remember that the formula contains \((n-1)\).

Correct answer: C. (88)

Here (a=-20), (d=6), so (a_{19}=-20+18\times6=88). In an increasing AP starting negative, add carefully at the end.

Correct answer: C. (48)

(d=\frac{32-16}{9-5}=4), so (a_{13}=32+4\times4=48). Equal position gaps give equal term gaps in an AP.

Correct answer: B. 3

The nth-term formula of an AP is \(a_n=a+(n-1)d\). Here, \(a=9\) and \(a_{21}=69\), so \(69=9+(21-1)d=9+20d\). Hence, \(20d=60\), giving \(d=3\). If 2 were used, the 21st term would be \(9+20\times2=49\), not 69. Exam tip: the nth term contains \((n-1)d\), not \(nd\).

Correct answer: B. 11th term

Here, the first term is \(a=45\) and the common difference is \(d=-5\). Therefore, \(a_n=a+(n-1)d=45-5(n-1)=50-5n\). For the first negative term, \(50-5n<0\), so \(n>10\). The smallest integer satisfying this is \(11\); hence, the 11th term is the first negative term. The 10th term is \(0\), which is not negative. Exam tip: For a “first” term question, choose the smallest positive integer that satisfies the inequality.

Correct answer: A. 83

Here, the first term is \(a=6\) and the common difference is \(d=7\). Thus, the terms increase by 7: \(\ldots, 76, 83, 90\). Therefore, the last term just less than \(90\) is \(83\). The numbers \(84\), \(85\), and \(86\) are not terms of this AP. Exam tip: To identify the term immediately before a given number, also verify the next AP term.