Finding the $n$th term of an AP

Use the general relationship between two terms of an arithmetic progression: a_n=a_r+(n-r)d. The known term is a_2=12, the target is a_7, and the common difference is 9. There are 7-2=5 equal steps from the second term to the seventh term. Hence a_7=12+5×9=12+45=57. Therefore option B is correct. The calculation must count the gaps between term positions, not the target index itself. Adding four or six differences would incorrectly give 48 or 66, while the other listed values are distractors based on such indexing or arithmetic errors.

For this AP, the first term is \(a=48\) and the common difference is \(d=42-48=-6\). The \(n\)th term is \(a_n=a+(n-1)d\). Hence, \(a_{11}=48+(11-1)(-6)=48-60=-12\). Therefore, \(-12\) is correct. A value such as \(-14\) may result from miscounting terms or using \(n\) instead of \((n-1)\). Exam tip: always use \((n-1)\) in the formula for the \(n\)th term.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{10}=2.5+(10-1)\times2.5=2.5+22.5=25\). Hence, the correct answer is \(25\). The value \(27.5\) would result from incorrectly using \(10d\) instead of \((10-1)d\). Exam tip: Always use \((n-1)\) in the formula for the \(n\)th term.

In this AP, the first term is \(a=7\) and the common difference is \(d=16-7=9\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{21}=7+(21-1)\times9=7+180=187\). Hence, 187 is correct. Getting 185 usually results from using an incorrect number of common differences. Exam tip: from the first term to the \(n\)th term, add the common difference \(n-1\) times.

The (12)th term is (4d) after the (8)th term, so (44+4(-2)=36). Add the negative difference carefully.

For this AP, the first term is \(a=12\) and the common difference is \(d=17-12=5\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{24}=12+(24-1)\times5=12+115=127\). Therefore, 127 is correct. A value such as 125 results from an incorrect count of the common differences. Exam tip: always use \((n-1)d\), not \(nd\), in the nth-term formula.

The first term is \(a=-15\) and the common difference is \(d=-9-(-15)=6\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{18}=-15+(18-1)\times6=-15+102=87\). Hence, \(87\) is correct. A close error such as \(85\) can result from incorrectly counting the number of common differences. Exam tip: always use \(n-1\) common differences to find the \(n\)th term of an AP.

The nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{13}=120+(13-1)(-9)=120-108=12\). Therefore, 12 is correct. Getting 21 is a common error caused by using the negative common difference \(-9\) with the wrong sign. Exam tip: always use \(n-1\) and keep a negative \(d\) in brackets while calculating.

The first term of the AP is 6 and the common difference is \(17-6=11\). Using \(a_n=a+(n-1)d\), \(a_{16}=6+(16-1)\times11=6+165=171\). Hence, 171 is correct. A value such as 169 would result from using an incorrect common difference or term count. Exam tip: for the \(n\)th term, add \(n-1\) common differences, not \(n\).

The first term is \(a=33\), and the common difference is \(d=29-33=-4\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{14}=33+(14-1)(-4)=33-52=-19\). Hence, \(-19\) is correct. The value \(-17\) results from adding \(-4\) only 12 times, which gives the 13th term. Exam tip: Always use \(n-1\), not \(n\), in the nth-term formula.

The nth term of an AP is \(a_n=a_1+(n-1)d\). Thus, \(a_5=22+(5-1)\times13=22+52=74\). Therefore, 74 is correct. The value 76 results from incorrectly adding \(5d\); the common difference is added only four times to reach the fifth term. Exam tip: always use \(n-1\), not \(n\), in the formula.

For this AP, the first term is \(a=9\) and the common difference is \(d=15-9=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{28}=9+(28-1)\times6=9+162=171\). Hence, 171 is correct. Getting 169 usually indicates an arithmetic error while adding \(27\times6\). Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

The (15)th term is (6d) after the (9)th term, so (70+18=88). This method is simple for nearby terms.

The first term is \(a=14\) and the common difference is \(d=14-14=0\). Hence, \(a_n=a+(n-1)d=14+(n-1)\times 0=14\). Therefore, the 40th term is 14. The value 560 is \(14\times 40\), but an AP term is not found by multiplying the first term by its position. Exam tip: if all terms of an AP are equal, its common difference is 0 and every term equals the first term.

Here (d=1), so (a_6=\frac{4}{5}+5=\frac{29}{5}). Convert the whole number to a fraction with the same denominator.

The standard form is \(a_n=a+(n-1)d\). Comparing it with \(12-3(n-1)\) gives \(a=12\) and \(d=-3\). The negative difference means each next term decreases by 3. Exam tip: carefully retain the sign of the coefficient of \((n-1)\).

Here (a=12) and (d=-3), so (a_{25}=12+24(-3)=-60). Keep the sign of (d) carefully in decreasing APs.

From (103=5+(n-1)7), (98=7(n-1)), so (n=15). When finding the term number, remember to add (1) at the end.

For an AP, the sixth term is \(a_6=a+5d\). Thus, \(23=a+5(5)=a+25\), so \(a=-2\). If the first term were 2, the sixth term would be \(2+25=27\), so that option is incorrect. Exam tip: in the \(n\)th-term formula, the coefficient of \(d\) is always \(n-1\).

From (88=3+(n-1)5), (85=5(n-1)), hence (n=18). In term-number questions, treat the given term as (a_n).