Finding the $n$th term of an AP

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{4m}-a_m=(4m-m)d=3m\times6\). Since \(156-48=108\), we get \(18m=108\), so \(m=6\). For instance, if \(m=7\), the difference would be \(18\times7=126\), not 108. Exam tip: When two AP terms are given, subtract them directly instead of first finding the first term.

For this AP, the first term is \(a=73\) and the common difference is \(d=65-73=-8\). Thus, \(a_n=73-8(n-1)\). Here \(a_{20}=-79\), which is greater than \((-80)\), while the next term \(a_{21}=-87\) is less than \((-80)\). Therefore, the first such term is \((-87)\). Exam tip: In a decreasing AP, check the two consecutive terms around the given boundary to identify the first required term.

For an AP, \(a_{18}-a_7=(18-7)d\). Thus, \((x+90)-(x+13)=11d\), so \(77=11d\) and \(d=7\). Now \(a_{32}=a_{18}+(32-18)d=x+90+14\times7=x+188\). Therefore, option B is correct. Choosing \(x+195\) would incorrectly use a gap of 15 terms, whereas the gap from the 18th to the 32nd term is 14. Exam tip: always subtract the term numbers to find the number of common differences between two terms.

Here, \(a=14\) and \(d=-3\). Thus \(a_n=a+(n-1)d=14-3(n-1)=17-3n\). In option B, putting \(n=1\) gives 11, not 14. Exam tip: always verify the first term.

The nth term of an AP is \(a_n=a+(n-1)d\). Substituting \(a=14\) and \(d=-3\) gives \(a_n=14-3(n-1)=17-3n\). A rule containing \(n^2\) does not produce a constant difference. Exam tip: identify \(a\) and \(d\) first.

The nth term of an AP is \(a_n=a+(n-1)d\). Here, \(a_n=4-3(n-1)=7-3n\), so A is correct. Option C has a positive common difference. Exam tip: check the signs of \(a\) and \(d\) first.

Here, the first term is 17 and the common difference is 10. Thus, \(a_n=17+10(n-1)=10n+7\). From \(10n+7>500\), we get \(n>49.3\), so the least integer value of \(n\) is 50. Therefore, \(a_{50}=507\), which is the first term greater than 500. The close distractor 497 is the preceding term, since \(a_{49}=497\). Exam tip: For “greater than”, use the next integer value after solving the inequality.

Since \(p=9\), we have \(a_9=42\) and \(a_{21}=150\). Thus, \(a_{21}-a_9=12d=150-42=108\), so \(d=9\). Now \(a_{40}=a_9+(40-9)d=42+31\times9=321\). The value 330 does not follow from the common difference \(d=9\) for the 40th term. Exam tip: When two terms with different indices are given, first use the difference of their indices to find \(d\).

For an AP, \(a_p-a_q=(p-q)d\). Since \(p\ne q\) and \(a_p=a_q\), we get \(d=0\), so every term is equal. Exam tip: equal terms at distinct indices imply zero common difference.

In an AP, the difference between consecutive terms is constant. For \(a_n=4-3n\), \(a_{n+1}-a_n=-3\), so it is an AP. For \(4-3n^2\), the difference depends on \(n\). Exam tip: check first differences.

Rewrite \(a_n=7-3n\) as \(4+(n-1)(-3)\), so the common difference is \(-3\). A negative difference makes the AP decreasing; \(3\) would give an increasing AP. Exam tip: identify \(d\) from the coefficient of \(n\).

The first term is \(a=9\), and the common difference is \(d=\frac{23}{2}-9=\frac{5}{2}\). The nth term is \(a_n=a+(n-1)d\). Therefore, \(a_{42}=9+41\times\frac{5}{2}=\frac{18+205}{2}=\frac{223}{2}\). \(\frac{225}{2}\) can result from an incorrect count of intervals or using the wrong value in place of \(n-1\). Exam tip: always use \(n-1\), not \(n\), in the AP nth-term formula.

For an AP, \(a_{37}-a_{15}=(37-15)d\). Thus, \(-154-0=22d\), so \(d=-7\). Now, using \(a_{15}=a_1+14d\), we get \(0=a_1+14(-7)\), hence \(a_1=98\). If 84 were chosen, the 15th term would not be 0. Exam tip: Find \(d\) first by subtracting the two given terms.

The nth term of an AP has the form \(a_n=a+(n-1)d\), so it is linear in \(n\). In \(n^2+1\), consecutive differences are 3, 5, 7, ..., not constant. \(5\) is still an AP with \(d=0\). Exam tip: a squared or cubed \(n\) usually rules out an AP.

In \(a_n=a+(n-1)d\), the coefficient of \(n\) is the common difference \(d\). Here it is \(-3\), so every next term decreases by 3. Note that 7 is not the first term. Exam tip: identify \(d\) from the coefficient of \(n\).

For an AP, \(a_n=a+(n-1)d\). Therefore, \(a_{22}-a_7=(22-7)d=15d\). From the given relation, \(15d=120\), so \(d=8\). If the common difference were 9, the difference would be \(15\times 9=135\), not 120. Exam tip: when two terms are given, subtract their indices and multiply the result by \(d\).

In option A, \(a_{n+1}-a_n=[7-3(n+1)]-(7-3n)=-3\), which is constant for every \(n\); hence it is an AP. In option B, the difference changes. Exam tip: always test consecutive-term differences.

From the 4th to the 9th term and from the 9th to the 14th term, the gap is 5 terms each. \(a_9-a_4=(9x-16)-(4x-1)=5x-15\), so \(a_{19}-a_{14}\) is also \(5x-15\). Hence, \(a_{19}=(14x-31)+(5x-15)=19x-46\). The option \(19x-44\) has an incorrect addition of the constant terms. Exam tip: In an AP, equal gaps between term numbers give equal differences between the corresponding terms.

Here, the first term is 29 and the common difference is 12. Therefore, the nth term is \(a_n=29+12(n-1)\). For a term less than 800, \(29+12(n-1)<800\), which gives \(n-1<64.25\). Thus, the greatest possible integer value is \(n-1=64\), and the term is \(29+12\times64=797\). The next term, 809, is greater than 800. Exam tip: To find the greatest AP term below a limit, solve the inequality and take the greatest valid integer value of n.

For an AP, \(a_5=a+4d\). Thus, \(18=a+4\times3\) gives \(a=6\), so \(a_n=6+(n-1)3=3n+3\). In option B, the fifth term would be 21. Exam tip: find \(a\) before writing the general term.

Let the first term be \(a\) and the common difference be \(d\). Then \(a_3+a_8=(a+2d)+(a+7d)=2a+9d=76\), while \(a_{13}=a+12d=92\). Multiplying the second equation by 2 gives \(2a+24d=184\). Subtracting the first equation gives \(15d=108\), so \(d=7.2\). Hence \(a=92-12(7.2)=5.6\), and \(a_{27}=a+26d=5.6+26(7.2)=192.8\). The nearby distractor 196 does not satisfy the given conditions. Exam tip: first convert every stated AP term or sum into equations in \(a\) and \(d\).

Putting \(n=1\) gives \(a_1=12-4=8\). The coefficient of \(n\) is the common difference, so \(d=-4\); hence the AP decreases. Do not mistake the constant 12 for the first term; substitute \(n=1\).

The first term is \(a=-11\), and the common difference is \(d=\frac{-9}{2}-(-11)=\frac{13}{2}\). Therefore, \(a_{27}=a+(27-1)d=-11+26\times\frac{13}{2}=-11+169=158\). Hence, 158 is correct. \(169\) represents only \(26d\); the first term \(-11\) must also be added. Exam tip: while finding the \(n\)th term, use \(a_n=a+(n-1)d\), not \(a+nd\).

For option A, \(a_{n+1}-a_n=[4(n+1)-7]-(4n-7)=4\), which is constant, so it is an AP. For \(n^2-7\), the difference \(2n+1\) varies. Exam tip: \(pn+q\) form gives an AP.

In an AP, \(a_{19}-a_8=(19-8)d=11d\). Since \(a_{19}=a_8+99\), we get \(11d=99\), so \(d=9\). Now, \(a_{52}=a_8+(52-8)d=46+44\times9=46+396=442\). Therefore, the correct answer is \(442\). An option such as \(436\) results from not adding the correct number of common differences. Exam tip: when two terms are given, first use the difference in their indices to find \(d\).