Finding the $n$th term of an AP

The first term is \(a=13\), and the common difference is \(d=\frac{39}{2}-13=\frac{13}{2}\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{46}=13+45\times\frac{13}{2}=\frac{26+585}{2}=\frac{611}{2}\). Hence, option B is correct. A nearby option such as \(\frac{599}{2}\) can result from using an incorrect number of differences. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

For an AP, \(a_{58}-a_{22}=(58-22)d\). Thus, \(-468-0=36d\), so \(d=-13\). Now, from \(a_{22}=a_1+21d\), \(0=a_1+21(-13)\), giving \(a_1=273\). Hence, 273 is the correct option. If 263 were used, the 22nd term would not be 0. Exam tip: First find \(d\) by using the difference between the given term numbers.

For \(a_n=4n-7\), \(a_{n+1}-a_n=[4(n+1)-7]-(4n-7)=4\), so the common difference is constant and it is an AP. The differences in the other options vary with \(n\). Exam tip: in \(pn+q\), the common difference is \(p\).

For this AP, the first term is 310 and the common difference is \(d=291-310=-19\). Thus, each successive term decreases by 19. The term after 82 is \(82-19=63\). Since 82 is greater than 75 while 63 is less than 75, 63 is the first term below 75. Although 44 and 25 are also less than 75, they occur after 63. Exam tip: In a decreasing AP, compare the terms immediately on either side of the given boundary.

For an AP, \(a_n=a+(n-1)d\). Therefore, \(a_{35}-a_{11}=(35-11)d=24d\). Given \(a_{35}-a_{11}=288\), we get \(24d=288\), so \(d=288/24=12\). The gap between the term numbers is 24; using an incorrect gap gives a wrong value. Exam tip: use \(a_m-a_n=(m-n)d\) directly in such questions.

For option B, \(a_{n+1}-a_n=[5-3(n+1)]-(5-3n)=-3\), which is constant for every \(n\). Hence it is an AP. In option A, the difference depends on \(n\). Exam tip: test \(a_{n+1}-a_n\) for constancy.

The term indices increase by 8 each time. Hence, \(a_{15}-a_7=(15x-48)-(7x-8)=8x-40\), and \(a_{23}-a_{15}=8x-40\). Therefore, the next equal-step difference is also \(8x-40\). Thus, \(a_{31}=a_{23}+(8x-40)=23x-88+8x-40=31x-128\). The option \(31x-124\) results from an incorrect calculation of the constant term by 4. Exam tip: Terms separated by equal index gaps in an AP have equal differences.

Here, the first term is 52 and the common difference is 21. Thus, \(a_n=52+21(n-1)=21n+31\). For a term not exceeding 2000, \(21n+31\le 2000\), giving \(n\le 93\). Therefore, \(a_{93}=1984\) is the greatest term in the given range. The next term is 2005, which exceeds 2000. Exam tip: To find the greatest AP term below a limit, use the greatest integer value of n satisfying the inequality.

For \(a_n=5n-3\), the consecutive difference is \(a_{n+1}-a_n=5\), which is constant; hence it is an AP. In \(n^2+1\), the difference changes. Exam tip: any form \(pn+q\) represents an AP.

Let the first term be \(a\) and the common difference be \(d\). Then \(a_6+a_{13}=(a+5d)+(a+12d)=2a+17d=211\), while \(a_{22}=a+21d=256\). Doubling the second equation and subtracting the first gives \(25d=301\), so \(d=\frac{301}{25}\). Hence \(a=256-21d=\frac{79}{25}\). Therefore, \(a_{44}=a+43d=\frac{79}{25}+43\cdot\frac{301}{25}=\frac{13022}{25}\). The value \(539\) does not follow because the given data do not give \(d=14\). Exam tip: write \(a_n=a+(n-1)d\) carefully, especially when converting term indices.

In an AP, \(a_n=a+(n-1)d\). Thus, \(a_{12}=a+11d\) and \(a_{22}=a+21d\); their average is \(a+16d=a_{17}\). In option B, the midpoint of the indices is 16.5, not 17. Exam tip: a term midway between two indices equals the average of those terms.

The common difference is −17/2 − (−19) = 21/2. Using aₙ = a₁ + (n − 1)d, a₃₉ = −19 + 38 × 21/2 = 380. Therefore, option B is correct.

Use the AP relation aₙ = aₘ + (n − m)d. Since a₁₄ = 51 and a₃₀ = 8a₁₄, we first calculate a₃₀ = 8 × 51 = 408. The index difference between 30 and 14 is 16, so a₃₀ − a₁₄ = 16d. Hence 408 − 51 = 357 = 16d. We do not need to simplify d fully, because the indices 46 and 30 also differ by 16. Therefore a₄₆ = a₃₀ + 16d = 408 + 357 = 765. Thus option B is correct. A common error is to multiply 51 by 8 again when finding a₄₆, or to use 46 − 14 without accounting for the already known term a₃₀. The equal index gaps provide the quickest reliable solution.

In an AP, the difference between two terms equals the difference in their indices multiplied by the common difference. Thus, \(a_{29}-a_{12}=17d=255\), so \(d=15\). Now, \(a_{72}=a_{12}+(72-12)d=86+60\times15=986\). Hence, 986 is correct. The nearby distractor 996 may result from an arithmetic error of 10. Exam tip: when using two known terms, always use the difference between their indices to find the common difference.

For \(a_n=5n-2\), \(a_{n+1}-a_n=[5(n+1)-2]-(5n-2)=5\), which is constant; hence it is an AP. For \(n^2+1\), the differences change. Exam tip: test consecutive-term differences.

Here, the first term is \(a=240\) and the common difference is \(d=-18\). Thus, \(a_n=240-18(n-1)\). Using \(a_n>-150\), we get \(240-18(n-1)>-150\), so \(n-1<\frac{65}{3}\). Therefore, the greatest possible integer is \(n=22\), and \(a_{22}=240-18(21)=-138\). The next term, \(a_{23}=-156\), is not greater than \(-150\). Exam tip: For a decreasing AP, solve the inequality and take the greatest valid integer value of \(n\).

Using aₖ = a₁ + (k − 1)d with k = 5n + 4 gives 555 = 23 + (5n + 3) × 14. Thus 532 = 14(5n + 3), so 5n + 3 = 38 and n = 7. Therefore, option B is correct.

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{8n}-a_{3n}=(8n-3n)d=5n\times24=120n\). Given \(810-210=600\), we get \(120n=600\), so \(n=5\). If \(n=6\), the term difference would be \(720\), not the given difference of 600. Exam tip: subtract the given terms to eliminate the first term quickly.

In an AP, consecutive terms have a constant difference, so its general term is linear in \(n\). For option C, \(a_{n+1}-a_n=2n+1\), which varies with \(n\). Exam tip: test the first difference.

The direct answer is option C, 402. In an arithmetic progression, equal increases or decreases occur between consecutive terms. The terms a20 and a52 have indices 20 and 52, and 36 is exactly halfway between them because 36−20=16 and 52−36=16. Therefore, the middle term is the average of the two end terms: a36=(a20+a52)/2=(178+626)/2=804/2=402. Option A, 382, is too small and does not equal this average. Option B, 392, is also below 402. Option C, 402, is correct because the required index is exactly midway between 20 and 52. Option D, 412, is too large. A useful exam cue is: in an AP, if the indices are equally spaced, the corresponding terms are also equally spaced, so the middle term is the average of the outer terms.

The difference between consecutive terms is \((w-4)-(w-15)=11\), so \(d=11\) and the first term is \(a=w-15\). Using \(a_n=a+(n-1)d\), we get \(677=(w-15)+36\times11=w+381\). Hence, \(w=296\). If \(w=300\), the 37th term would be 681, not 677. Exam tip: for the \(n\)th term of an AP, use \(n-1\) common differences.

Given \(a_n=17n-11\), \(a_{7k}=17(7k)-11=119k-11\) and \(a_{3k}=17(3k)-11=51k-11\). Hence, \(a_{7k}-a_{3k}=(119k-11)-(51k-11)=68k\). Since \(68k=408\), we get \(k=6\). The constant term \(-11\) cancels when the two terms are subtracted. Exam tip: substitute each index separately in \(a_n\) before taking their difference.

The nth-term form of an AP is \(a_n=a+(n-1)d\). Here, \(5-4n=1+(n-1)(-4)\), so the first term is \(1\) and the common difference is \(-4\). A negative difference still forms an AP. Exam tip: put \(n=1\) to find the first term quickly.

The positive multiples form the AP \(41, 82, 123, \ldots\). The last term must be less than 5000. Since \(41\times121=4961\), and the next multiple is \(41\times122=5002\), which is greater than 5000, the last term is \(4961\). Also, \(4920\) is not a multiple of 41. Exam tip: divide by the given number, take the integer part, and verify the resulting multiple against the limit.

The first term of the AP is \(a=1517\), and its common difference is \(d=1554-1517=37\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Hence, \(a_{31}=1517+(31-1)\times37=1517+1110=2627\). Therefore, option B is correct. Using \(2609\) results from an incorrect term count or from not using \((n-1)\). Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.