Finding the $n$th term of an AP

In an AP, the difference between two terms equals the difference in their indices multiplied by the common difference. Thus, \(a_{29}-a_{12}=17d=255\), so \(d=15\). Now, \(a_{72}=a_{12}+(72-12)d=86+60\times15=986\). Hence, 986 is correct. The nearby distractor 996 may result from an arithmetic error of 10. Exam tip: when using two known terms, always use the difference between their indices to find the common difference.

For \(a_n=5n-2\), \(a_{n+1}-a_n=[5(n+1)-2]-(5n-2)=5\), which is constant; hence it is an AP. For \(n^2+1\), the differences change. Exam tip: test consecutive-term differences.

Here, the first term is \(a=240\) and the common difference is \(d=-18\). Thus, \(a_n=240-18(n-1)\). Using \(a_n>-150\), we get \(240-18(n-1)>-150\), so \(n-1<\frac{65}{3}\). Therefore, the greatest possible integer is \(n=22\), and \(a_{22}=240-18(21)=-138\). The next term, \(a_{23}=-156\), is not greater than \(-150\). Exam tip: For a decreasing AP, solve the inequality and take the greatest valid integer value of \(n\).

From (260=1250+(n-1)(-55)), (990=55(n-1)), so (n=19). In a decreasing situation, take (d) as negative.

From (514=64+(n-1)15), (450=15(n-1)), so (n=31). Treat the daily count as a term, not a total sum.

(527=23+(5n+3)14), so (504=70n+42), which does not give an integer option. Match the index and options carefully.

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{8n}-a_{3n}=(8n-3n)d=5n\times24=120n\). Given \(810-210=600\), we get \(120n=600\), so \(n=5\). If \(n=6\), the term difference would be \(720\), not the given difference of 600. Exam tip: subtract the given terms to eliminate the first term quickly.

In an AP, consecutive terms have a constant difference, so its general term is linear in \(n\). For option C, \(a_{n+1}-a_n=2n+1\), which varies with \(n\). Exam tip: test the first difference.

(a_{36}) is equally spaced between (a_{20}) and (a_{52}). Therefore (a_{36}=\frac{178+626}{2}=402).

The difference between consecutive terms is \((w-4)-(w-15)=11\), so \(d=11\) and the first term is \(a=w-15\). Using \(a_n=a+(n-1)d\), we get \(677=(w-15)+36\times11=w+381\). Hence, \(w=296\). If \(w=300\), the 37th term would be 681, not 677. Exam tip: for the \(n\)th term of an AP, use \(n-1\) common differences.

Given \(a_n=17n-11\), \(a_{7k}=17(7k)-11=119k-11\) and \(a_{3k}=17(3k)-11=51k-11\). Hence, \(a_{7k}-a_{3k}=(119k-11)-(51k-11)=68k\). Since \(68k=408\), we get \(k=6\). The constant term \(-11\) cancels when the two terms are subtracted. Exam tip: substitute each index separately in \(a_n\) before taking their difference.

The nth-term form of an AP is \(a_n=a+(n-1)d\). Here, \(5-4n=1+(n-1)(-4)\), so the first term is \(1\) and the common difference is \(-4\). A negative difference still forms an AP. Exam tip: put \(n=1\) to find the first term quickly.

The positive multiples form the AP \(41, 82, 123, \ldots\). The last term must be less than 5000. Since \(41\times121=4961\), and the next multiple is \(41\times122=5002\), which is greater than 5000, the last term is \(4961\). Also, \(4920\) is not a multiple of 41. Exam tip: divide by the given number, take the integer part, and verify the resulting multiple against the limit.

The first term of the AP is \(a=1517\), and its common difference is \(d=1554-1517=37\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Hence, \(a_{31}=1517+(31-1)\times37=1517+1110=2627\). Therefore, option B is correct. Using \(2609\) results from an incorrect term count or from not using \((n-1)\). Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

Let the first term be \(a\) and common difference be \(d\). Then \(a_{10}+a_{30}=2a+38d=500\) and \(a_{18}+a_{38}=2a+54d=820\). Subtracting gives \(16d=320\), so \(d=20\). From \(2a+38(20)=500\), we get \(a=-130\). Hence, \(a_{58}=a+57d=-130+57(20)=1010\). The nearby option 1030 can result from incorrectly using \(58d\); remember that \(a_{58}=a+57d\). Exam tip: use \(a_n=a+(n-1)d\).

For \(a_n=5-2n\), \(a_{n+1}-a_n=[5-2(n+1)]-(5-2n)=-2\), which is constant for every n. Hence it is an AP. The differences for \(3n^2-2\) are not constant. Exam tip: in \(a_n=pn+q\), the common difference is \(p\).

Here, the first term is \(a=75\) and the common difference is \(d=98-75=23\). The \(n\)th term is \(a_n=75+23(n-1)\). Using \(75+23(n-1)<2500\), we get \(n-1<\frac{2425}{23}\approx105.43\). Hence, the greatest integer value of \(n\) is \(106\). Therefore, 106 terms are less than 2500. For \(n=107\), the term is \(2513\), which is not less than 2500. Exam tip: retain the strict inequality \((<)\) when the question says “less than.”

The standard AP form is \(a_n=a+(n-1)d\). Here, \(a_1=7(1)-3=4\), and each successive term increases by 7, so \(d=7\). The \(-3\) is only a constant term, not the common difference. Exam tip: find \(a_1\) and \(a_2\), then subtract.

(a_{16}+a_{25}=(a_7+9d)+(a_7+18d)=116+27d=638), so (d=\frac{58}{3}) and (a_{49}=58+42d=870).

Here, the first term is \(a=-92\) and the common difference is \(d=23\). Thus, \(a_n=a+(n-1)d=-92+23(n-1)=23n-115\). From \(23n-115>900\), we get \(n>44.13\), so the smallest integer value is \(n=45\). Hence, \(a_{45}=920\), which is the first term greater than 900. The close distractor 897 is the 44th term, so it is still less than 900. Exam tip: for the first term greater than a number, solve the inequality and take the next integer value of \(n\).