Finding the $n$th term of an AP

In an AP, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_{21}-a_9=12d=132\), so \(d=11\). Now, \(a_{57}=a_9+(57-9)d=54+48\times11=582\). Therefore, option C is correct. The value 570 does not include the full increase of 48 common differences. Exam tip: When moving from one known term to another, use the difference between their term numbers.

The nth term of an AP is linear: \(a+(n-1)d\). For \(5n-3\), \(a_{n+1}-a_n=5\), a constant difference. For \(n^2+1\), the difference is \(2n+1\), which varies. Exam tip: check consecutive-term differences.

Here, the first term is \(a=121\) and the common difference is \(d=-12\). Thus, \(a_n=121-12(n-1)\). We get \(a_{17}=-71\), and the next term is \(a_{18}=-83\). Since \(-71>-75\) but \(-83<-75\), the last term greater than \(-75\) is \((-71)\). \((-83)\) is a close distractor, but it is less than \(-75\). Exam tip: In a decreasing AP, check the two consecutive terms on either side of the given limit.

From (280=760+(n-1)(-32)), (480=32(n-1)), so (n=16). In a decreasing situation, take (d) as negative.

From (276=42+(n-1)9), (234=9(n-1)), so (n=27). Treat the daily count as a term, not a total sum.

From (275=11+(3n+3)8), (264=24n+24), so (n=10). Subtract (1) from the index to get the multiplier of (d).

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{6n}-a_{2n}=(6n-2n)d=4n\times14=56n\). Given \(319-95=224\), we get \(56n=224\), so \(n=4\). Therefore, option 4 is correct. Exam tip: subtract the given terms to eliminate the first term; there is no need to find it separately.

For option A, \(a_{n+1}-a_n=\frac{5-2(n+1)}{3}-\frac{5-2n}{3}=-\frac{2}{3}\), which is constant; hence it is an AP. In B, the difference changes. Exam tip: an AP has an nth term linear in \(n\).

(a_{26}) is equally spaced between (a_{14}) and (a_{38}). Therefore (a_{26}=\frac{92+284}{2}=188).

The difference between consecutive terms is \(8\), so the first term is \(a=u-9\) and the common difference is \(d=8\). Using \(a_n=a+(n-1)d\), \(295=(u-9)+28\times8=u+215\). Hence, \(u=80\). If \(u=78\), the 29th term would be \(293\), so it is not correct. Exam tip: for the \(n\)th term, use \(n-1\) common differences, not \(n\).

Given \(a_n=13n-6\), we have \(a_{5k}=13(5k)-6=65k-6\) and \(a_{2k}=13(2k)-6=26k-6\). Hence, \(a_{5k}-a_{2k}=(65k-6)-(26k-6)=39k\). Now \(39k=234\), so \(k=234/39=6\). If \(k=5\), the difference would be only \(195\), so it is not correct. Exam tip: substitute each index separately in the term formula before subtracting.

For \(a_n=7n-4\), \(a_{n+1}-a_n=7\), a constant non-zero value for every \(n\); hence it is an AP. Terms involving \(n^2\), \(2^n\), or \(1/n\) do not have constant consecutive differences. Exam tip: a linear form \(pn+q\) always represents an AP.

The positive multiples of 29 form the AP \(29, 58, 87, \ldots\), whose \(n\)th term is \(29n\). For the last term, \(29n<2500\). Since \(2500\div29\approx86.2\), the greatest integer value of \(n\) is 86. Hence the last term is \(29\times86=2494\). Although \(2523=29\times87\), it is greater than 2500. Exam tip: for “less than”, use the integer part of the quotient and verify the product.

The first term is \(a=920\), and the common difference is \(d=943-920=23\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{27}=920+(27-1)\times23=920+598=1518\). The value 1541 results from incorrectly using \(n\) instead of \((n-1)\). Exam tip: there are always \(n-1\) common differences between the first term and the \(n\)th term.

For an AP, \(a_n=a+(n-1)d\). Thus, \(a_7+a_{21}=2a+26d=224\) and \(a_{13}+a_{27}=2a+38d=368\). Subtracting the equations gives \(12d=144\), so \(d=12\). Then \(2a+26(12)=224\) gives \(a=-44\). Hence, \(a_{41}=a+40d=-44+40(12)=436\). Option 560 results from not accounting for the first term \(a\). Exam tip: in AP questions involving two sums of terms, subtract the equations first to find \(d\) quickly.

(a_{11}=90) and (5d=62), so (d=\frac{62}{5}). (a_{31}=90+20\cdot\frac{62}{5}=338), so the options do not contain the correct answer.

Here, the first term is \(a=42\) and the common difference is \(d=13\). The \(n\)th term is \(a_n=42+13(n-1)\). From \(42+13(n-1)<1000\), we get \(13n<971\), so \(n<74.69\). Hence, there are \(74\) terms less than 1000. The \(75\)th term is \(1017\), which is not less than 1000. Exam tip: after solving an inequality for the number of terms, take the greatest integer satisfying it.

The \(n\)th term is obtained by adding the common difference \(d\), \((n-1)\) times to the first term \(a\); hence \(a_n=a+(n-1)d\). Using \(a+nd\) adds one extra \(d\) even for the first term. Exam tip: put \(n=1\); the result must be \(a\).

(a_{12}+a_{19}=(a_5+7d)+(a_5+14d)=62+21d=242), so (d=\frac{60}{7}) and (a_{33}=31+28d=271). Check the options carefully.

Here, the first term is \(a=-47\) and the common difference is \(d=16\). Thus, \(a_n=-47+16(n-1)=16n-63\). For \(a_n>500\), \(16n-63>500\), so \(n>35.1875\). The smallest integer value is \(n=36\), giving \(a_{36}=513\). Option 497 is not greater than 500. Exam tip: For “greater than,” use the next integer term satisfying the inequality.