Finding the $n$th term of an AP

Here, the first term is \(a=3\) and the common difference is \(d=8-3=5\). Thus, \(a_n=a+(n-1)d=3+5(n-1)=5n-2\). Option B gives 7 when \(n=1\), not 3. Exam tip: identify \(a\) and \(d\) first.

For an AP, \(a_n=a_3+(n-3)d\). Thus, \(a_7=14+4d\) and \(a_{11}=14+8d\). Hence, \((14+4d)+(14+8d)=100\), so \(28+12d=100\), giving \(d=6\). Now, \(a_{19}=a_3+16d=14+16\times6=110\). Therefore, 110 is the correct answer. The value 108 may result from incorrectly taking the number of steps from the third term to the nineteenth term as 15. Exam tip: when starting from \(a_3\), use the index difference \((n-3)\).

Here, the first term is \(a=-25\) and the common difference is \(d=11\). Thus, \(a_n=-25+11(n-1)=11n-36\). For \(a_n>200\), \(11n-36>200\), so \(n>21.45\). The smallest integer value is \(n=22\), giving \(a_{22}=206\). The value \(195\) is the preceding term, \(a_{21}\), so it is not greater than 200. Exam tip: For a “first term greater than” question, take the smallest integer satisfying the inequality.

Given \(a_n=5n+q\), we have \(a_{4n}=5(4n)+q=20n+q\). Hence, \(a_{4n}-a_n=(20n+q)-(5n+q)=15n\). Since \(15n=135\), \(n=9\). Therefore, option C is correct. The constant \(q\) cancels because it occurs in both terms. Exam tip: substitute the given indices into the term formula first, then simplify the difference.

\(a_{n+1}-a_n=[p(n+1)+q]-(pn+q)=p\), which is constant for every \(n\). Hence the sequence is an AP with common difference \(p\), not \(q\). Exam tip: test an AP by subtracting consecutive terms.

(d=\frac{148-57}{24-11}=7), so (a_{41}=148+17\times7=267). First find the common difference and move from the nearer term.

(a_{14}=74) and (8d=51), so (d=\frac{51}{8}). From the nearer term (a_{30}=74+16\cdot\frac{51}{8}=176), so check options carefully.

Given \(a_n=5n+s\), we have \(a_{18}=5\times18+s=90+s\). Since \(a_{18}=112\), \(s=22\). Therefore, \(a_{46}=5\times46+22=230+22=252\). Option \(247\) would result from incorrectly taking 27 gaps; actually, \(46-18=28\). Exam tip: You may also use \(a_{46}=a_{18}+(46-18)d\), where \(d=5\).

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.