Finding the $n$th term of an AP

In an AP, \(a_{17}-a_7=(17-7)d=10d\). Hence, \(10d=60\), so \(d=6\). Now, \(a_{43}=a_7+(43-7)d=33+36\times6=249\). Therefore, 249 is correct. Getting 255 would result from an incorrect term gap or common difference. Exam tip: the number of common differences between two terms equals the difference between their subscripts.

For \(a_n=3n^2-2\), \(a_1=1,a_2=10,a_3=25\), so the consecutive differences are \(9\) and \(15\), not constant. An AP has a linear nth-term form \(pn+q\). Exam tip: check whether first differences are equal.

The first term is 85 and the common difference is \(d=-8\). \((-35)\) is itself a term of the AP because \(85-8\times15=-35\). The term immediately before it is \((-35)+8=-27\), which is greater than \((-35)\). Hence, \((-27)\) is the correct answer. \((-35)\) is not greater than the given limit; it is the limit itself. Exam tip: In a decreasing AP, add \(|d|\) to get the term just before a known term.

The water quantities form an arithmetic progression with first term \(a=360\) and common difference \(d=-24\). In the \(n\)th hour, the quantity is \(a_n=360+(n-1)(-24)\). From \(360-24(n-1)=96\), we get \(24(n-1)=264\), so \(n-1=11\) and \(n=12\). Therefore, the water will be 96 litres in the 12th hour. In the 11th hour it would be 120 litres, making it a close but incorrect option. Exam tip: use a negative common difference when a quantity decreases.

The daily study times form an arithmetic progression with first term \(a=27\) and common difference \(d=6\). The study time on the \(n\)th day is \(a_n=a+(n-1)d\). Thus, \(165=27+(n-1)\times6\), so \(138=6(n-1)\) and \(n-1=23\). Hence, \(n=24\), meaning the student will study for 165 minutes on the 24th day. On the 23rd day, the time would be only \(159\) minutes. Exam tip: While finding a term number, remember that the AP formula contains \(n-1\).

The general term of an AP is \(a_r=a_1+(r-1)d\). Here, putting \(r=2n+3\), we get \(159=5+(2n+2)\times7\). Thus, \(154=14n+14\), so \(14n=140\) and \(n=10\). If 9 is used, the term would be 145, not 159. Exam tip: Substitute the complete expression \(2n+3\) for the index \(r\) in \(a_r\).

For an AP, \(a_{4n}-a_n=(4n-n)d=3nd\). Thus, \(140-32=3n\times6\), so \(108=18n\). Hence, \(n=6\). Note that the difference between the term indices is \(3n\), not \(4n\). Exam tip: When two terms are given, subtract their indices first and multiply the result by \(d\).

Given \(a_6=22\), we get \(a_{14}=3(22)+4=70\). In an AP, \(a_{14}-a_6=(14-6)d=8d\). Hence, \(70-22=8d\), so \(d=6\). Now \(a_{22}=a_6+(22-6)d=22+16\times6=118\). Therefore, the correct answer is \(118\). The option \(114\) results from not applying the correct difference in term numbers. Exam tip: use \(a_m-a_n=(m-n)d\) when two terms of an AP are known.

The difference between the first two terms is \((c+1)-(c-5)=6\), so \(d=6\) and the first term is \(a=c-5\). The 21st term is \(a_{21}=a+20d\). Hence, \(139=(c-5)+20\times6=c+115\), which gives \(c=24\). If \(c=26\), the 21st term would be 141, so it is not correct. Exam tip: for the \(n\)th term of an AP, use \(a+(n-1)d\).

Given \(a_n=9n+2\), we have \(a_{3k}=9(3k)+2=27k+2\) and \(a_k=9k+2\). Hence, \(a_{3k}-a_k=(27k+2)-(9k+2)=18k\). Now \(18k=144\), so \(k=8\). Therefore, option C is correct. For the close distractor 9, the difference would be \(18\times9=162\), not 144. Exam tip: Write the terms for both indices separately before subtracting them.

The nth term of an AP is \(a_n=a+(n-1)d\). With \(a=-3\) and \(d=4\), \(a_n=-3+4(n-1)=4n-7\). In option B, substituting \(n=1\) gives first term 1. Exam tip: verify both the first term and coefficient of \(n\).

For \(a_n=7-3n\), \(a_{n+1}-a_n=[7-3(n+1)]-(7-3n)=-3\), which is constant for every \(n\). Hence it is an AP. In option B, the difference is \(2n-2\), so it changes. Exam tip: test whether consecutive-term differences are constant.

The positive multiples of 19 form the AP \(19, 38, 57, \ldots\). For the last term, \(19n<1500\). Since \(1500\div19\approx78.94\), the greatest integer value of \(n\) is 78. Thus, the last term is \(19\times78=1482\). Although \(1501=19\times79\), it is greater than 1500. Exam tip: for “less than,” take the integer part of the quotient and verify the multiple.

The first term is \(a=510\), and the common difference is \(d=527-510=17\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{19}=510+(19-1)\times17=510+306=816\). Hence, 816 is correct. The value 833 results from incorrectly counting 20 differences instead of the 18 differences from the first term to the 19th term. Exam tip: always use \((n-1)\), not \(n\), in the nth-term formula.

Let the first term be \(a\) and the common difference be \(d\). Then \(a_5+a_{15}=2a+18d=100\) and \(a_9+a_{19}=2a+26d=164\). Subtracting the first equation from the second gives \(8d=64\), so \(d=8\). Using \(2a+18\times8=100\), we get \(a=-22\). Hence, \(a_{27}=a+26d=-22+26\times8=186\). The value 210 does not follow from the correct common difference. Exam tip: subtract equations formed from sums of AP terms to find \(d\) quickly.

For an AP, \(a_n=a+(n-1)d=dn+(a-d)\). Comparing with \(7n-3\) gives \(d=7\) and \(a-d=-3\), so \(a=4\). Option B wrongly treats −3 as the first term. Exam tip: in linear nth-term form, the coefficient of \(n\) is the common difference.

Here, the first term is 24 and the common difference is 9. Therefore, \(a_n=24+9(n-1)\). Using the condition \(a_n<400\), we get \(24+9(n-1)<400\), or \(n<42.78\). Hence, the greatest possible integer value is \(n=42\), so 42 terms are less than 400. For \(n=43\), the term is 402, which is not less than 400. Exam tip: For “less than” questions, retain the strict inequality \(<\) and select the greatest valid integer.

Here, the first term is \(a=3\) and the common difference is \(d=8-3=5\). Thus, \(a_n=a+(n-1)d=3+5(n-1)=5n-2\). Option B gives 7 when \(n=1\), not 3. Exam tip: identify \(a\) and \(d\) first.

For an AP, \(a_n=a_3+(n-3)d\). Thus, \(a_7=14+4d\) and \(a_{11}=14+8d\). Hence, \((14+4d)+(14+8d)=100\), so \(28+12d=100\), giving \(d=6\). Now, \(a_{19}=a_3+16d=14+16\times6=110\). Therefore, 110 is the correct answer. The value 108 may result from incorrectly taking the number of steps from the third term to the nineteenth term as 15. Exam tip: when starting from \(a_3\), use the index difference \((n-3)\).

Here, the first term is \(a=-25\) and the common difference is \(d=11\). Thus, \(a_n=-25+11(n-1)=11n-36\). For \(a_n>200\), \(11n-36>200\), so \(n>21.45\). The smallest integer value is \(n=22\), giving \(a_{22}=206\). The value \(195\) is the preceding term, \(a_{21}\), so it is not greater than 200. Exam tip: For a “first term greater than” question, take the smallest integer satisfying the inequality.

Given \(a_n=5n+q\), we have \(a_{4n}=5(4n)+q=20n+q\). Hence, \(a_{4n}-a_n=(20n+q)-(5n+q)=15n\). Since \(15n=135\), \(n=9\). Therefore, option C is correct. The constant \(q\) cancels because it occurs in both terms. Exam tip: substitute the given indices into the term formula first, then simplify the difference.

\(a_{n+1}-a_n=[p(n+1)+q]-(pn+q)=p\), which is constant for every \(n\). Hence the sequence is an AP with common difference \(p\), not \(q\). Exam tip: test an AP by subtracting consecutive terms.

The common difference is d = (148 − 57)/(24 − 11) = 7. From a₂₄ to a₄₁ there are 17 steps, so a₄₁ = 148 + 17 × 7 = 267.

First, a₁₄ = 3 × 23 + 5 = 74. Since a₁₄ − a₆ = 74 − 23 = 51 across 8 steps, the common difference is d = 51/8. From a₁₄ to a₃₀ there are 16 steps, so a₃₀ = 74 + 16 × (51/8) = 176.

Given \(a_n=5n+s\), we have \(a_{18}=5\times18+s=90+s\). Since \(a_{18}=112\), \(s=22\). Therefore, \(a_{46}=5\times46+22=230+22=252\). Option \(247\) would result from incorrectly taking 27 gaps; actually, \(46-18=28\). Exam tip: You may also use \(a_{46}=a_{18}+(46-18)d\), where \(d=5\).