Finding the $n$th term of an AP

The first term is \(a=-7\), and the common difference is \(d=\frac{-3}{2}-(-7)=\frac{11}{2}\). Using \(a_n=a+(n-1)d\), we get \(a_{18}=-7+17\times\frac{11}{2}=\frac{-14+187}{2}=\frac{173}{2}\). Hence, \(\frac{173}{2}\) is correct. The option \(\frac{175}{2}\) can result from a small error in counting terms or arithmetic. Exam tip: for an AP involving fractions, write terms with a common denominator first to check the common difference accurately.

Given \(a_8=28\), \(a_{20}=3a_8=3\times28=84\). In an AP, \(a_{20}-a_8=(20-8)d=12d\). Hence, \(84-28=12d\), so \(d=\frac{14}{3}\). Now \(a_{26}=a_{20}+(26-20)d=84+6\times\frac{14}{3}=84+28=112\). Therefore, 112 is correct. Choosing 106 would mean adding the increase over six terms incorrectly. Exam tip: when two distant terms are given, use the difference of their indices to find \(d\).

In an AP, \(a_{15}-a_5=(15-5)d=10d\). Since \(a_{15}=a_5+50\), we get \(10d=50\), so \(d=5\). Now, \(a_{31}=a_5+(31-5)d=24+26\times5=154\). Hence, 154 is correct. Although 158 is close, it does not follow from the given common difference \(d=5\). Exam tip: When comparing two terms, use the difference between their term numbers.

Given \(a_n=an+b\), we have \(a_{14}-a_4=10a\). Thus, \(71-21=10a\), so \(a=5\). From the 14th term to the 30th term, there are 16 term-intervals. Therefore, \(a_{30}=a_{14}+16\times5=71+80=151\). The value 147 would result from using an incorrect interval count. Exam tip: subtract two given terms to eliminate the constant \(b\) and find the common difference/slope directly.

The first term is 64 and the common difference is \(d=57-64=-7\). Thus, each successive term decreases by 7: \(1,-6,-13,-20\). Here, \((-13)\) is greater than \((-20)\), while the next term is \((-20)\), which is not greater than itself. Therefore, the answer is \((-13)\). Exam tip: For “greater than”, do not include the boundary value itself.

The number of questions solved each day forms an AP with first term \(a=18\) and common difference \(d=4\). The number solved on the \(n\)th day is \(a_n=a+(n-1)d\). Thus, \(102=18+(n-1)4\), so \(84=4(n-1)\) and \(n-1=21\). Hence, \(n=22\). On the 21st day, the student solves 98 questions, so 102 questions are solved on the 22nd day. Exam tip: identify whether the question asks for a term of the AP or for the total sum.

The governing concept is the nth-term formula for an arithmetic progression: a_k = a₁ + (k − 1)d. Here the term index is k = 2n + 1, so the number of common-difference steps from the first term is k − 1 = 2n, not 2n + 1. Substituting the data gives 123 = 3 + (2n)(4). Therefore, 120 = 8n, and dividing by 8 gives n = 15. Hence option B is correct. A frequent error is to multiply d by the full index 2n + 1; that would incorrectly count one extra step. Substitution of n = 15 gives k = 31 and a₃₁ = 3 + 30 × 4 = 123, confirming the result. The other options do not satisfy the stated AP equation.

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{3n}-a_n=(3n-n)d=2nd\). Substituting the given values gives \(81-21=2n\times5\), so \(60=10n\). Therefore, \(n=6\). If \(n=5\), the difference would be \(2\times5\times5=50\), not 60. Exam tip: subtract the two term equations first; the first term \(a\) cancels automatically.

Use the arithmetic-progression relation a_n=a_k+(n−k)d. First evaluate the twelfth term from the given condition: a_12=5a_4−8=5(17)−8=85−8=77. Therefore a_12−a_4=77−17=60. The index gap from the fourth term to the twelfth term is 8, so 8d=60 and d=7.5. From a_4 to a_20 there are 20−4=16 equal steps. Hence a_20=a_4+16d=17+16(7.5)=17+120=137. Thus option A is correct. The other values would require a different common difference and would not preserve the relation between the fourth and twelfth terms. A fractional common difference is acceptable in an AP.

The common difference between consecutive terms is \(5\), so the first term is \(a=b-3\) and \(d=5\). The \(18\)th term is \(a_{18}=a+17d\). Thus, \(94=(b-3)+17\times5=b+82\), which gives \(b=12\). If \(b=11\), the \(18\)th term would be \(93\), so it is not correct. Exam tip: use \(n-1\) common differences when finding the \(n\)th term.

Given a_n = 7n - 4, we get a_{2k} = 7(2k) - 4 = 14k - 4 and a_k = 7k - 4. Hence, a_{2k} - a_k = (14k - 4) - (7k - 4) = 7k. So, 7k = 84 gives k = 12. If k were 11, the difference would be 77, not 84. Exam tip: when the subscript is 2k, substitute the complete expression 2k for n.

The nth term of an AP is \(a_n=a_1+(n-1)d\). Thus, \(a_n=52+(n-1)(-4)=56-4n\). For a negative term, \(56-4n<0\), so \(n>14\). The smallest integer satisfying this is \(n=15\), and hence \(a_{15}=-4\) is the first negative term. At \(n=14\), the term is \(0\), which is not negative. Exam tip: when the question asks for the “first” term, choose the smallest integer greater than the value obtained from the inequality.

Given \(a_n=11n+c\), use \(a_5=72\): \(72=11\times5+c\), so \(c=17\). Now \(a_{5r}=512\) gives \(512=11(5r)+17=55r+17\). Hence \(55r=495\), so \(r=9\). If \(r=10\), the term would be \(567\), not \(512\). Exam tip: first find the constant \(c\) from the given term, then substitute the required index in the formula.

The AP of positive multiples of 17 is \(17, 34, 51, \ldots\). Its last term has the form \(17n\), with \(17n<1000\). Since \(1000\div17\approx58.82\), the greatest integer value of \(n\) is 58. Therefore, the last term is \(17\times58=986\). Although \(1003=17\times59\), it is not less than 1000. Exam tip: For “less than” limits, take the integer part of the quotient and verify by multiplication.

The first term of the AP is \(a=260\), and the common difference is \(d=273-260=13\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_{22}=260+(22-1)\times13=260+273=533\). Therefore, 533 is correct. The number 546 is the next, or 23rd, term. Exam tip: use \(n-1\), not \(n\), in the formula for the \(n\)th term.

Let the first term be a and the common difference be d. Using a_r = a + (r − 1)d, the first condition becomes (a + 3d) + (a + 11d) = 2a + 14d = 64. The second becomes (a + 7d) + (a + 15d) = 2a + 22d = 112. Subtracting gives 8d = 48, so d = 6. Substituting into 2a + 14d = 64 gives 2a + 84 = 64 and hence a = −10. Therefore a₂₀ = a + 19d = −10 + 19×6 = 104. Thus option C is correct. The originally listed alternatives did not contain 104, so the option set has been corrected; 78, 84, and 112 do not satisfy the given equations.

For an AP written as \(a_n=A+Bn\), the difference \(a_{n+1}-a_n=B\). In option A, the coefficient of \(n\) is \(-\frac{5}{2}\). Option B has a positive difference, while C has difference \(-5\). Exam tip: inspect the coefficient of \(n\).

Here, the first term is 18 and the common difference is 7. The nth term is 18+7(n-1). From 18+7(n-1)<200, we get 7n<189, so n<27. Hence, the greatest integer value of n is 26, and there are 26 terms less than 200. The 27th term is 200, so it is not included. Exam tip: For “less than”, do not count a term equal to the given limit.

Subtracting the given terms gives \(a_{25}-a_{10}=15d=149-59=90\). Hence, \(d=6\). Now \(a_{40}=a_{25}+15d=149+15\times6=239\). Therefore, the correct answer is 239. Choosing 245 would incorrectly use 16 common differences; there are only 15 differences from the 25th term to the 40th term. Exam tip: To find \(d\) quickly, subtract two known terms and divide by the difference in their term numbers.

In an AP, the difference between consecutive terms is constant. Since \(a_5=a_2+3d\) and \(a_8=a_2+6d\), we get \(9+3d+9+6d=72\). Thus, \(18+9d=72\), so \(d=6\). Now \(a_{14}=a_2+12d=9+12\times6=81\). Hence, 81 is correct. A value such as 73 may result from incorrectly counting the number of common differences between the terms. Exam tip: from \(a_r\) to \(a_s\), the change is \((s-r)d\).

Here, the first term is \(a=-12\) and the common difference is \(d=7\). Thus, \(a_n=-12+7(n-1)=7n-19\). For \(a_n>100\), we get \(7n-19>100\), so \(n>17\). The smallest integer value is \(n=18\), and \(a_{18}=107\). While 100 is the 17th term, the next AP term is 107. Exam tip: for “greater than,” solve the inequality and take the smallest valid integer value of \(n\).

Given \(a_n=4n+q\), we get \(a_{3n}=4(3n)+q=12n+q\). Hence, \(a_{3n}-a_n=(12n+q)-(4n+q)=8n\). Since \(8n=96\), \(n=12\). The constant \(q\) cancels when the two terms are subtracted, so it does not affect the value of \(n\). Exam tip: Substitute \(3n\) for \(n\) carefully before finding the difference.

The common difference is \(d=\frac{a_{18}-a_6}{18-6}=\frac{-70-2}{12}=-6\). Now, \(a_{30}=a_{18}+(30-18)d=-70+12(-6)=-142\). Hence, the correct answer is \(-142\). Choosing \(-136\) would result from using an incorrect difference in term numbers. Exam tip: when two terms are given, always use the difference between their subscripts.

The common difference is d = (116 − 44) ÷ (21 − 9) = 72 ÷ 12 = 6. From the 21st term to the 38th term there are 17 steps, so a₃₈ = 116 + 17 × 6 = 218.

Use the AP relation a_n = a_m + (n − m)d. First calculate a₁₃ from the given condition: a₁₃ = 2a₅ + 9 = 2(25) + 9 = 59. From the fifth term to the thirteenth term there are 8 steps, so a₁₃ = a₅ + 8d. Thus 59 = 25 + 8d, giving 8d = 34 and d = 17/4. From a₅ to a₂₉ there are 29 − 5 = 24 steps. Therefore a₂₉ = a₅ + 24d = 25 + 24(17/4) = 25 + 102 = 127. Hence option A is correct. The other choices arise from using an incorrect step count, such as 23 or 25, or from an arithmetic mistake. The fractional common difference is valid because an AP need not have an integer difference.