Finding the $n$th term of an AP

(a_{20}=209) and (12d=172), so (d=\frac{43}{3}). (a_{32}=209+12\cdot\frac{43}{3}=381), which is not in the options.

(a_{32}) is equally spaced between (a_{18}) and (a_{46}). Therefore (a_{32}=\frac{135+443}{2}=289).

The difference between consecutive terms is \((v-3)-(v-12)=9\), so \(d=9\) and the first term is \(a=v-12\). The nth term is \(a_n=a+(n-1)d\). Hence, \(438=(v-12)+30\times9=v+258\), giving \(v=180\). If \(v=176\), the 31st term would be \(434\), so it is not correct. Exam tip: In an AP, multiply the common difference by \(n-1\), not by \(n\).

Given \(a_n=15n-8\), we get \(a_{6k}=15(6k)-8=90k-8\) and \(a_{2k}=15(2k)-8=30k-8\). Therefore, \(a_{6k}-a_{2k}=(90k-8)-(30k-8)=60k\). Using \(60k=300\), we obtain \(k=5\). If \(k=4\), the difference would be only \(240\), so it is not correct. Exam tip: when subtracting two terms of this form, the identical constant terms cancel out.

In \(a_n=a+(n-1)d\), the coefficient of \(n\) is the common difference \(d\). Here it is \(-4\), so \(d=-4\) and the terms decrease. As a quick check, putting \(n=1\) gives the first term \(3\), not \(7\).

The sequence of positive multiples is \(37,74,111,\ldots\), whose \(n\)th term is \(37n\). For the last term, \(37n<3600\). Now \(37\times97=3589\), whereas \(37\times98=3626>3600\). Hence, 3589 is the greatest multiple of 37 less than 3600. Although 3626 is the next multiple, it exceeds the given limit. Exam tip: divide by the common difference, take the integer part, and multiply back to find the required multiple.

The first term of the AP is \(a=1209\), and the common difference is \(d=1240-1209=31\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{29}=1209+(29-1)\times31=1209+868=2077\). Hence, 2077 is correct. Using \(29\times31\) would incorrectly add 29 gaps; from the first term to the 29th term, there are only 28 gaps. Exam tip: always use \((n-1)\) in the nth-term formula.

For an AP, \(a_n=a+(n-1)d\). Hence, \(a_8+a_{24}=2a+30d=320\) and \(a_{14}+a_{30}=2a+42d=512\). Subtracting the equations gives \(12d=192\), so \(d=16\). Then \(2a+30(16)=320\), giving \(a=-80\). Therefore, \(a_{46}=a+45d=-80+45(16)=640\). A common mistake is to use \(46d\) instead of \((46-1)d\) in the term formula. Exam tip: when sums of AP terms are given, form both linear equations and subtract them first.

(a_{13}=132) and (6d=96), so (d=16). (a_{37}=36+30\times16=516), so the correct answer is not in the options.

Here, the first term is 56 and the common difference is 17. Therefore, \(a_n=56+17(n-1)\). Using \(a_n<1500\), we get \(56+17(n-1)<1500\), so \(n-1<84.94\ldots\). Hence, the greatest integer value of n is 85. In fact, \(a_{85}=1484\), whereas \(a_{86}=1501\), which is not less than 1500. Exam tip: For “less than,” do not include a term equal to the limiting value.

The nth term of an AP is \(a_n=a+(n-1)d\), so it must be linear in n. For option B, \(a_{n+1}-a_n=2n-2\), which changes with n; hence the common difference is not constant. Exam tip: check whether the first difference is fixed.

(a_{14}+a_{22}=(a_6+8d)+(a_6+16d)=84+24d=420), so (d=14). (a_{38}=42+32d=490), which is not in the options.

Here, the first term is \(a=-68\) and the common difference is \(d=19\). Thus, \(a_n=-68+19(n-1)\). For \(a_n>700\), \(-68+19(n-1)>700\), giving \(n>41.42\). The smallest integer value is \(n=42\). Therefore, \(a_{42}=711\), so 711 is the first term greater than 700. Although 692 is the nearest preceding term, it is less than 700. Exam tip: for the “first term greater than” condition, solve the inequality and take the smallest valid integer \(n\).

Given \(a_n=10n+q\), we get \(a_{7n}=70n+q\) and \(a_{3n}=30n+q\). Hence, \(a_{7n}-a_{3n}=(70n+q)-(30n+q)=40n\). Therefore, \(40n=520\), so \(n=13\). The constant \(q\) cancels because it occurs in both terms. Exam tip: for terms such as \(a_{7n}\) and \(a_{3n}\), directly substitute \(7n\) and \(3n\) respectively in the term formula.

Comparing with \(a_n=a+(n-1)d\), we get \(a=4\) and \(d=-7\). A negative common difference means each next term is 7 less, so the AP decreases. Exam tip: the coefficient of \((n-1)\) is the common difference.

(d=\frac{205-77}{29-13}=8) so (a_{47}=205+18\times8=349). Moving from the nearer known term is faster.

(a_{4p}-a_p=3pd=132) so (33p=132) and (p=4). In index questions first find the position gap.

Expanding option A gives \(a_n=n^2+5n+4-n^2=5n+4\). It has the form \(an+b\), so its common difference is the constant \(5\). Option B contains an \(n^2\) term, so successive differences cannot remain constant. Exam tip: an AP’s nth term is linear in \(n\).

In this AP, the first term is 260 and the common difference is \(d=-15\). \((-100)\) is itself a term of the AP, since \(260-15\times 24=-100\). As each successive term decreases by 15, the term after \((-100)\) is \((-100)-15=-115\). Therefore, the first term less than \((-100)\) is \((-115)\). Although \((-130)\) is also less than \((-100)\), it occurs later. Exam tip: when asked for the “first” term beyond a boundary, check the term immediately after the boundary term.

In an AP, \(a_{n+1}-a_n\) must be constant for every \(n\). For option A, \([5(n+1)-2]-(5n-2)=5\). In B, the difference changes with \(n\). Exam tip: look for the linear form \(pn+q\).