Finding the $n$th term of an AP

(a_{18}) is equally spaced between (a_9) and (a_{27}). Therefore (a_{18}=\frac{46+154}{2}=100).

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.

The first sum gives (2a+14d=64) and the second gives (2a+22d=112). From (8d=48), (d=6) and (a_{20}=84).

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.

(d=\frac{116-44}{21-9}=6), so (a_{38}=116+17\times6=218). First find (d), then move forward from the nearer term.

\(a_{13}=59\) and \(8d=34\), so \(d=\frac{17}{4}\). \(a_{29}=25+24\cdot\frac{17}{4}=127\).

Given \(a_n=4n+r\). Substituting \(n=15\), \(83=4\times15+r=60+r\), so \(r=23\). Hence, \(a_{41}=4\times41+23=164+23=187\). Therefore, option C is correct. The value \(183\) would result from using an incorrect value of \(r\). Exam tip: first find the unknown constant from the given term, then substitute the required value of \(n\).

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \,\(a_{3m}-a_m=(3m-m)d=2md\). Substituting the given values gives \,\(94-34=2m\times5\), so \,\(60=10m\). Hence, \,\(m=6\). If 5 were used, the difference between the terms would be only 50, so it is incorrect. Exam tip: In such questions, subtract the given terms instead of first finding the first term.

The first term is \(a=58\) and the common difference is \(d=-7\). Hence, \(a_n=58+(n-1)(-7)=65-7n\). For \(a_n<-60\), we need \(65-7n<-60\), so \(n>125/7\). The smallest integer satisfying this is \(n=18\). Therefore, \(a_{18}=65-7(18)=-61\). The previous term is \(a_{17}=-54\), which is not less than \(-60\). Although \(-68\) is also less than \(-60\), it occurs later in the AP. Exam tip: when asked for the first term meeting an inequality, use the smallest integer \(n\) that satisfies it.