Finding the $n$th term of an AP

Here (a=-10), (d=4), so (a_{18}=-10+17\times4=58). The formula remains the same even when terms move from negative to positive.

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{10}=8+(10-1)\times 9=8+81=89\). Therefore, 89 is correct. The value 90 can result from an incorrect use of the common difference. Exam tip: for the nth term, use \((n-1)\), not \(n\).

Here (a=5), (d=7), so (a_{17}=5+16\times7=117). Check the difference using adjacent terms.

Here (a=2), (d=7), so (a_{22}=2+21\times7=149). The (22)nd term adds (21) common differences.

Here (d=-3), so (a_{15}=30+14(-3)=-12). Watch the sign carefully in a decreasing AP.

The first row is term 1, so the increase of 3 occurs only 11 times before term 12: \(18+11\times3=51\). Using \(18+12\times3\) counts one extra increase. In exams, always use \(n-1\).

The first term of this AP is 4 and the common difference is 0, since \(4-4=0\) for consecutive terms. Using \(a_n=a+(n-1)d\), \(a_{50}=4+49\times0=4\). Choosing 50 confuses the term number with the value of the term. Exam tip: find the common difference first; if \(d=0\), every term equals the first term.

In this AP, the first term is \(a=6\) and the common difference is \(d=10-6=4\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{31}=6+(31-1)\times4=6+120=126\). Choosing 122 would correspond to adding only 29 differences, so it is close to the 30th term. Exam tip: use \(n-1\), not \(n\), when finding the \(n\)th term.

In this AP, the first term is \(a=9\) and the common difference is \(d=5-9=-4\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_8=9+(8-1)(-4)=9-28=-19\). Hence, option B is correct. \(-15\) is the seventh term, since \(a_7=9+6(-4)=-15\). Exam tip: use \(n-1\) in the nth-term formula and put a negative common difference in brackets.

The nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{13}=3+(13-1)\times10=3+120=123\). Hence, 123 is correct. The value 120 is only \(12d\); the first term, 3, must also be added. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

Here (a=11), (d=5), so (a_9=11+8\times5=51). One less than the term number is added as differences.

Given \(a_n=2n+3\), substitute \(n=18\) to find the 18th term: \(a_{18}=2(18)+3=36+3=39\). Therefore, 39 is correct. The option 41 can result from an arithmetic error while multiplying or adding. Exam tip: substitute the term number carefully into the formula before simplifying.

The general term is \(a_n=5n-1\). To find the 20th term, substitute \(n=20\): \(a_{20}=5(20)-1=100-1=99\). Therefore, 99 is correct. The value 101 would result from adding 1 instead of subtracting it. Exam tip: In an \(n\)th-term question, first substitute the required term number for \(n\).

Given \(a_n=7n+2\), substitute \(n=11\) to get \(a_{11}=7\times11+2=77+2=79\). Therefore, 79 is correct. The close distractor 77 results from forgetting to add \(+2\). In exams, substitute the term number carefully for \(n\).

To find the sixth term, substitute \(n=6\) in the given general term: \(a_6=30-3(6)=30-18=12\). Hence, the correct answer is 12. The value 18 is only \(3\times6\); it must still be subtracted from 30. Exam tip: Always substitute the term number carefully for \(n\) in \(a_n\).

The nth term of an AP is \(a_n=a_1+(n-1)d\). Therefore, \(a_{10}=13+(10-1)\times 8=13+72=85\). Hence, 85 is correct. The value 93 results from incorrectly adding \(10d\) instead of \((10-1)d\). Exam tip: for the nth term, always use \((n-1)\) common differences.

Here (a=18), (d=4), so (a_{24}=18+23\times4=110). Use the formula directly even for a larger term.

The nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_5=40+(5-1)(-6)=40-24=16\). Hence, the correct answer is 16. Getting 20 would mean subtracting the common difference only three times. Exam tip: always use \((n-1)\) common differences for the nth term.

Here (a=12), (d=3), so (a_{15}=12+14\times3=54). In word problems, treat rows as terms.

Here (a=5), (d=2), so (a_{12}=5+11\times2=27). Treat the day number as (n) and apply the formula.