Finding the $n$th term of an AP

In this AP, the first term is \(a=18\) and the common difference is \(d=23-18=5\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_8=18+(8-1)\times5=18+35=53\). Hence, \(53\) is correct. Choosing \(58\) would add one extra common difference, making it the 9th term. Exam tip: for the \(n\)th term, add only \(n-1\) common differences.

The first term is \(a=-5\), and the common difference is \(d=-2-(-5)=3\). Hence, \(a_{20}=a+(20-1)d=-5+19\times3=52\). Therefore, 52 is correct. Choosing 49 means counting only 18 intervals, but there are 19 intervals from the first term to the 20th term. Exam tip: Use \(a_n=a+(n-1)d\) and check \(n-1\) before calculating.

The (10)th term is (3d) after the (7)th term, so (31+3\times4=43). The difference method is quick for nearby terms.

The (8)th term is (3d) after the (5)th term, so (22+18=40). Counting the gap between terms is an easy method.

For this AP, the first term is \(a=40\) and the common difference is \(d=36-40=-4\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{15}=40+(15-1)(-4)=40-56=-16\). Hence, option \(\text{C}\) is correct. \(-20\) would be obtained for the 16th term, not the 15th term. Exam tip: for the \(n\)th term, use the common difference \((n-1)\) times.

The first term is \(a=2\) and the common difference is \(d=9-2=7\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{13}=2+(13-1)\times7=2+84=86\). Hence, \(86\) is correct. \(84\) results if the first term \(2\) is mistakenly not added. Exam tip: always use \((n-1)\) for the \(n\)th term of an AP.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Thus, \(a_9=1+(9-1)\times11=1+88=89\). Therefore, 89 is correct. A common error is choosing 88 by forgetting to include the first term; only 8 common differences are added to reach the 9th term. Exam tip: always use \((n-1)\), not \(n\), in the nth-term formula.

For this AP, the first term is \(a=16\) and the common difference is \(d=12-16=-4\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{10}=16+(10-1)(-4)=16-36=-20\). Hence, \(-20\) is correct. A value such as \(-18\) may result from an incorrect calculation or common difference. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

Here (a=5) and (d=5), so (a_{30}=5+29\times5=150). The formula works quickly for APs with equal multiples.

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Here, \(a=-2\), \(d=6\), and \(n=12\). Thus, \(a_{12}=-2+(12-1)\times6=-2+66=64\). Therefore, 64 is the correct answer. The nearby option 66 is only \(11\times6\); it does not include the first term \(-2\). Exam tip: use \((n-1)\), not \(n\), in the nth-term formula.

In this AP, the first term is \(a=21\) and the common difference is \(d=19-21=-2\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{16}=21+(16-1)(-2)=21-30=-9\). \(-7\) is the next term, i.e. the \(17\)th term. Exam tip: for the \(n\)th term, use \(n-1\) differences, not \(n\).

The (9)th term is (5d) after the (4)th term, so (18+15=33). Moving forward from the given term is quick.

The first term is \(a=12\) and the common difference is \(d=20-12=8\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{14}=12+(14-1)\times8=12+104=116\). Hence, 116 is correct. A value such as 114 results from an error in calculation or in finding the common difference. Exam tip: for the \(n\)th term, add \((n-1)d\), not \(nd\).

Here (a=\frac{3}{2}) and (d=1), so (a_8=\frac{17}{2}). Watch the denominator carefully in fractional terms.

The (7)th term is (4d) after the (3)rd term, so (14+20=34). Count the gap between terms correctly.

The first term is \(a=24\), and the common difference is \(d=21-24=-3\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{19}=24+(19-1)(-3)=24-54=-30\). Hence, \(-30\) is correct. A value such as \(-28\) can result from using an incorrect term number or common difference. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

In an AP, the common difference is added \(n-1\) times from the first term to reach the \(n\)th term, so \(a_n=a+(n-1)d\). Option B adds one extra \(d\). Exam tip: substitute \(n=1\); the result must be \(a\).

The first term of this AP is 17 and the common difference is 7. Thus, \(a_6=a+(6-1)d=17+5\times7=52\). Choosing 55 would add one extra difference instead of five. Exam tip: always use \(a_n=a+(n-1)d\) for the \(n\)th term.

Here (d=5), so (a_{11}=-8+10\times5=42). The formula remains the same even with a negative start.

The nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_9=60+(9-1)(-6)=60-48=12\). Therefore, 12 is the correct answer. You may get 18 if you incorrectly treat the negative common difference \(-6\) as positive. Exam tip: when \(d\) is negative, the terms decrease.