Finding the $n$th term of an AP

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.

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.

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.

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 \((n-1)\).

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.

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.

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 \(n-1\).

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.

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.

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}}).

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

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.

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.

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, \(n-m\).

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

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 \(n\).

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.

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.

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.

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 \((n-1)\).

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

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

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 \(nd\).

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.

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.