Finding the $n$th term of an AP

The first term is \(a=-9\). The common difference is \(d=\frac{-7}{2}-(-9)=\frac{11}{2}\). Therefore, \(a_{23}=a+(23-1)d=-9+22\times\frac{11}{2}=-9+121=112\). Hence, the correct answer is 112. \(\frac{225}{2}\) may result from not subtracting \(-9\) correctly at the final step. Exam tip: use \(n-1\), not \(n\), in the formula for the \(n\)th term.

The difference is \(a_{n+1}-a_n=[p(n+1)+q]-(pn+q)=p\), which is constant. Hence it is an AP; \(q\) affects the first term, not the common difference. Exam tip: subtract consecutive terms to identify 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.

(a_{22}) is equally spaced between (a_{10}) and (a_{34}). Therefore (a_{22}=\frac{54+174}{2}=114).

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.