Finding the $n$th term of an AP

First, a₁₂ = 3 × 19 + 4 = 61. Since a₁₂ − a₅ = 7d, we get 61 − 19 = 7d, so d = 6. Therefore, a₂₆ = a₅ + 21d = 19 + 21 × 6 = 145.

For \(a_n=7n-3\), \(a_{n+1}-a_n=[7(n+1)-3]-(7n-3)=7\), a constant difference; hence it is an AP. In \(n^2+7\), the differences vary. Exam tip: a linear expression in n represents an AP.

Substituting n = 8 gives 101 = 9 × 8 + c, so c = 29. Then a₅ᵣ = 45r + 29 = 344, so 45r = 315 and r = 7.

For option A, \(a_{n+1}-a_n=[5(n+1)-2]-(5n-2)=5\), which is constant for every \(n\); hence it is an AP. In option B, the difference changes. Exam tip: test consecutive differences first.

From the \(p\)th term to the \(n\)th term, the common difference is added \(n-p\) times, so \(a_n=a_p+(n-p)d\). Option C reverses the difference. Exam tip: count gaps between term positions, not the positions themselves.

In an AP, the difference between consecutive terms is constant. For \(a_n=7-3n\), \(a_{n+1}-a_n=-3\), so it is an AP. The \(n^2\) rule does not give a fixed difference. Exam tip: check whether the nth term is linear in n.

The common difference is 31/2 − 11 = 9/2. Using a_n = a_1 + (n − 1)d, we get a_44 = 11 + 43(9/2) = 409/2. Therefore, option A is correct.

For an AP, \(a_n=a_1+(n-1)d\). Subtracting the given terms gives \(a_{50}-a_{20}=30d=-270\), so \(d=-9\). Now \(a_{20}=a_1+19d=0\), hence \(a_1=-19(-9)=171\). If 180 were chosen, the 20th term would be 9, not 0. Exam tip: subtract two given terms first to eliminate \(a_1\) and find \(d\) quickly.

The consecutive-term difference is \(a_{n+1}-a_n=[p(n+1)+q]-(pn+q)=p\), which is constant. Hence it is an AP for all constant \(p,q\), including \(p=0\), when it is a constant AP. Exam tip: check consecutive differences.

The first term is 221 and the common difference is \(d=204-221=-17\). The terms decrease as \(\ldots,68,51,34,17,0\). Since 51 is greater than 50 and the very next term, 34, is less than 50, the required first term is 34. Although 17 is also less than 50, it is not the first such term. Exam tip: In a decreasing AP, check the consecutive terms immediately on either side of the given limit.

In an AP, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_{31}-a_9=(31-9)d=22d\). Given \(a_{31}-a_9=198\), we get \(22d=198\), so \(d=9\). Hence, 9 is the correct option. If \(d=8\), the difference would be \(22\times8=176\), not 198. Exam tip: Use \(a_m-a_n=(m-n)d\) for direct questions of this type.

The common difference is d = (172 − 52)/(23 − 8) = 8. From a_n = 52 + (n − 8)8, setting a_n = 292 gives n = 38. Since the index is 8r, 8r = 38, so r = 19/4. Therefore, option B is correct.

The term indices increase by 7 each time. \(a_{13}-a_6=(13x-33)-(6x-5)=7x-28\), and \(a_{20}-a_{13}=7x-28\). Therefore, the same difference is added from \(a_{20}\) to \(a_{27}\): \(a_{27}=20x-61+7x-28=27x-89\). Hence, option C is correct. Exam tip: For equally spaced terms of an AP, the differences between the terms are equal.

Here, the first term is 43 and the common difference is 17. The nth term is \(a_n=43+17(n-1)\). For the greatest term less than 1300, use \(43+17(n-1)<1300\), which gives \(n\leq 74\). Therefore, \(a_{74}=43+17\times73=1284\). The number 1299 is not a term of the AP, while 1301 is greater than 1300. Exam tip: To find the largest AP term within a limit, write the nth-term formula and solve the relevant inequality.

In an AP, \(a_{n+1}-a_n\) must be constant. For option A, \((12-3(n+1))-(12-3n)=-3\), so every next term decreases by 3. Option B increases instead. Exam tip: check the coefficient of \(n\) in a linear term formula.

For an AP, \(a_n=a+(n-1)d\). Hence, \(a_5+a_{11}=2a+14d=138\), so \(a+7d=69\). Also, \(a_{18}=a+17d=165\). Subtracting these equations gives \(10d=96\), so \(d=9.6\) and \(a=1.8\). Therefore, \(a_{36}=a+35d=1.8+35(9.6)=337.8\). Option 327 is incorrect because it does not follow from the common difference determined by the given conditions. Exam tip: Form equations from the given terms or sums of terms and subtract them to find \(d\) quickly.

First, a₁₄ = 2(43) + 17 = 103. Since a₁₄ − a₉ = 5d, the common difference is d = (103 − 43)/5 = 12. Therefore, a₃₄ = a₉ + 25d = 43 + 25(12) = 343.

In the standard form \(a_n=a+(n-1)d\), \(7-2n=5+(n-1)(-2)\). Hence, the first term is 5 and the common difference is −2. A negative \(d\) indicates a decreasing AP. Exam tip: verify the first term by putting \(n=1\).

In an AP, the difference between two terms equals the difference of their positions multiplied by the common difference. Thus, \(a_{25}-a_{10}=15d=180\), so \(d=12\). Now, \(a_{65}=a_{10}+(65-10)d=70+55\times12=730\). Hence, 730 is correct. The close distractor 710 may result from incorrectly counting the difference in term positions instead of using \(65-10\). Exam tip: use \(a_m-a_n=(m-n)d\) when two terms are given; finding the first term is unnecessary.

The general term of an AP is \(a_n=a+(n-1)d\), which is linear in \(n\). Since \(n^2+1\) contains a squared term, its consecutive differences are not constant. Exam tip: a quadratic term usually indicates that the sequence is not an AP.

Here, the first term is \(a=156\) and the common difference is \(d=-14\). The \(n\)th term is \(a_n=156+(n-1)(-14)=170-14n\). For a term greater than \((-120)\), \(170-14n>-120\), so \(n<20.71\). Thus the greatest integral value is \(n=20\), giving \(a_{20}=-110\). The next term, \(a_{21}=-124\), is not greater than \((-120)\). Exam tip: In a decreasing AP, always check the next term after finding the candidate last term.

The number of questions solved each day forms an AP with first term \(a=55\) and common difference \(d=12\). The number solved on the \(n\)th day is \(a_n=a+(n-1)d\). Thus, \(367=55+(n-1)\times12\), giving \(312=12(n-1)\), so \(n-1=26\) and \(n=27\). Therefore, the student will solve 367 questions on the 27th day. The closest distractor, the 26th day, would give only \(355\) questions. Exam tip: Since the question asks for one day's count, use \(a_n\), not the sum formula \(S_n\).

Using aₖ = a₁ + (k − 1)d with k = 4n + 3 gives a₍₄ₙ₊₃₎ = 17 + (4n + 2)10 = 40n + 37. Thus 40n + 37 = 397, so 40n = 360 and n = 9.

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{7n}-a_{3n}=(7n-3n)d=4n\times16\). Hence \(530-146=64n\), so \(384=64n\), giving \(n=6\). If 5 were used, the difference between the terms would be \(320\), not \(384\). Exam tip: subtract the two given terms directly; there is no need to find the first term.

First, a₂₀ = 6(37) − 13 = 209. From a₈ to a₂₀ there are 12 common differences, so 12d = 209 − 37 = 172. From a₂₀ to a₃₂ there are also 12 common differences; hence a₃₂ = 209 + 172 = 381.