Finding the $n$th term of an AP

(a_{10}=62) and (6d=48), so (d=8). Then (a_{22}=62+12\times8=158); if options mismatch, recheck the question data.

For \(a_n=5n-3\), \(a_{n+1}-a_n=5\), which is constant for every \(n\); hence it is an AP. For \(n^2+1\), the difference is \(2n+1\), which changes. Exam tip: test consecutive-term differences.

From (61=42+c), (c=19). (313=28r+19) gives (r=\frac{294}{28}=10.5), so no integer option is correct.

The terms are of the form (19+11(n-1)). The first term greater than (700) is (701) because the previous term is (690).

In an AP with an odd number of terms, the middle term is the arithmetic mean of the first and last terms. Hence, first term + last term = 2 × middle term. Exam tip: first check whether the number of terms is odd.

The standard AP form is \(a_n=a+(n-1)d\). Here, \(a_1=7-3(1)=4\), and each successive term changes by \(-3\); hence \(d=-3\). Exam tip: substitute \(n=1\) to verify the first term.

\(a_{n+1}-a_n=[5-3(n+1)]-(5-3n)=-3\), which is constant for every \(n\). Hence it is an AP; the negative difference makes it decreasing. Exam tip: the coefficient of \(n\) gives the common difference.

The first term is \(a=8\), and the common difference is \(d=\frac{25}{2}-8=\frac{9}{2}\). Therefore, \(a_{38}=a+(38-1)d=8+37\times\frac{9}{2}=\frac{16+333}{2}=\frac{349}{2}\). Hence, option B is correct. Choosing \(\frac{341}{2}\) results from an error in counting the number of differences or calculating \(d\). Exam tip: for the \(n\)th term, always use \(a_n=a+(n-1)d\), not \(a+nd\).

For an AP, \(a_{42}-a_{18}=(42-18)d\). Thus, \(-192-0=24d\), so \(d=-8\). Using \(a_{18}=a_1+17d\), we get \(0=a_1+17(-8)\), hence \(a_1=136\). If \(128\) were used, the 18th term would not be 0. Exam tip: Find \(d\) first by subtracting the two given terms.

An AP has nth term \(a_n=a+(n-1)d\). In \(20-3n\), the coefficient of n is \(-3\), so consecutive terms differ by \(-3\). For \(20-3n^2\), the difference is not constant. Exam tip: check the coefficient of n in a linear expression.

The common difference is \(d=154-169=-15\). The terms around the boundary are \(49, 34, 19,\ldots\). Since \(49\) is not less than \(40\), but the very next term \(34\) is less than \(40\), the first required term is \(34\). Exam tip: In a decreasing AP, write the consecutive terms around the given boundary to identify the first qualifying term.

In an AP, the difference between two terms equals the difference of their positions multiplied by the common difference: \(a_{27}-a_8=(27-8)d=19d\). Given \(a_{27}-a_8=171\), we get \(19d=171\), so \(d=9\). Hence, option C is correct. If the common difference were 8, the term difference would be \(19\times8=152\), not 171. Exam tip: use \(a_m-a_n=(m-n)d\) to solve such questions quickly.

(d=\frac{122-38}{12}=7). From (206=38+(7r-7)7), (7r=31), so no integer option is correct.

The terms are six positions apart, so their differences must be equal. \(a_{11}-a_5=(11x-20)-(5x-2)=6x-18\), and \(a_{17}-a_{11}=6x-18\). Hence, \(a_{23}=a_{17}+(6x-18)=(17x-38)+(6x-18)=23x-56\). Therefore, option C is correct. Exam tip: When equally spaced AP terms are given, use the equal difference between those terms to find the next one.

Here, the first term is \(a=31\) and the common difference is \(d=14\). Thus, \(a_n=31+14(n-1)\). To find the greatest term below \(1000\), use \(31+14(n-1)<1000\), which gives \(n-1\leq69\). Therefore, the term is \(31+14\times69=997\). The next term is \(1011\), which exceeds \(1000\). Exam tip: for the greatest AP term below a limit, use an inequality and take the greatest possible integer value of \(n-1\).

The common difference is \(a_{n+1}-a_n\). Here, \(a_{n+1}=p(n+1)+q=pn+p+q\), so the difference is \(p\). The constant \(q\) only shifts every term. Exam tip: identify the coefficient of \(n\).

Let the first term be \(a\) and the common difference be \(d\). Then \(a_4+a_9=(a+3d)+(a+8d)=2a+11d=95\), while \(a_{16}=a+15d=137\). Doubling the second equation and subtracting the first gives \(19d=179\), so \(d=\frac{179}{19}\). Therefore, \(a_{31}=a_{16}+15d=137+15\left(\frac{179}{19}\right)=\frac{5288}{19}\). The value \(263\) would require \(d=9\), which does not satisfy both given conditions. Exam tip: Use \(a_{31}=a_{16}+15d\) to calculate a distant term efficiently.

(a_{12}=96) and (4d=59), so (d=\frac{59}{4}). (a_{28}=37+20\cdot\frac{59}{4}=332), so none of the given options is correct.

Here (d=\frac{15}{2}). (a_{31}=-13+30\cdot\frac{15}{2}=212), so the options should be rechecked.

For an AP, \(a_n=a+(n-1)d\). If \(d=0\), then \(a_n=a\), so every term is constant. Setting only \(a=0\) does not make terms constant. Exam tip: the coefficient of \(n\) is \(d\).