Finding the $n$th term of an AP

In this AP, the first term is \(a=25\) and the common difference is \(d=31-25=6\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{10}=25+(10-1)\times6=25+54=79\). Therefore, 79 is correct. The value 73 would result from adding only 8 common differences, but the 10th term is reached after 9 differences. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

For this AP, the first term is \(a=55\) and the common difference is \(d=52-55=-3\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{20}=55+(20-1)(-3)=55-57=-2\). Hence, the correct answer is \(-2\). An answer such as \(-4\) may result from incorrectly using \(20d\); the correct multiplier is \(n-1=19\). Exam tip: in the nth-term formula of an AP, the common difference is added \((n-1)\) times.

The nth term of an AP is \(a_n=a+(n-1)d\). Substituting \(n=15\) gives \(a_{15}=a+14d\). The expression \(a+15d\) adds one extra common difference. Exam tip: always check the \(n-1\) factor.

In this AP, the first term is \(a=100\) and the common difference is \(d=97-100=-3\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{25}=100+(25-1)(-3)=100-72=28\). Hence, 28 is correct. Getting 30 may result from using an incorrect number of common differences instead of 24. Exam tip: for the \(n\)th term, always use \(n-1\) differences.

In an AP, there are 5 steps from the third term to the eighth term. Thus, \(a_8=a_3+(8-3)d=16+5\times7=51\). Therefore, the correct answer is 51. Choosing 49 would mean that the common difference has not been added the correct number of times. Exam tip: subtract the term numbers to find the number of common-difference steps between two terms.

Here (d=10), so (a_{12}=13+11\times10=123). The (12)th term includes (11) common differences.

The first term is \(a=-20\), and the common difference is \(d=-15-(-20)=5\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{15}=-20+(15-1)\times5=-20+70=50\). Hence, 50 is correct. The value 45 may result from incorrectly using 13 instead of \(n-1\). Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

The nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_6=18+(6-1)(-7)=18-35=-17\). Hence, \(-17\) is correct. An option such as \(-15\) may result from an error in multiplication or subtraction. Exam tip: When \(d\) is negative, keep it in brackets while substituting in the formula.

For this AP, the first term is \(a=11\) and the common difference is \(d=19-11=8\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{19}=11+(19-1)\times 8=11+144=155\). Hence, \(155\) is correct. \(153\) may result from an error in counting terms or multiplication. Exam tip: always use \((n-1)\), not \(n\), in the formula for the \(n\)th term.

In this AP, the first term is \(a=1.5\) and the common difference is \(d=3.0-1.5=1.5\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_8=1.5+(8-1)\times1.5=12.0\). Hence, \(12.0\) is correct. \(11.5\) is not correct because each successive term increases by \(1.5\). Exam tip: identify \(a\) and \(d\) before applying the \(a_n\) formula.

In an AP, to find a later term from a known term, multiply the common difference by the difference in term numbers. Thus, \(a_{13}=a_6+(13-6)d=38+7\times5=73\). Therefore, 73 is correct. Choosing 68 would add only 6 common differences, but there are 7 gaps from the 6th term to the 13th term. Exam tip: always subtract the term numbers to count the required common differences.

For this AP, the first term is \(a=64\) and the common difference is \(d=60-64=-4\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{18}=64+(18-1)(-4)=64-68=-4\). Hence, \(-4\) is correct. The close distractor \(-2\) is incorrect because adding \(-4\) 17 times decreases 64 by \(68\). Exam tip: Always use \(n-1\), not \(n\), in the nth-term formula.

The first term is \(a=5\) and the common difference is \(d=14-5=9\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{14}=5+(14-1)\times 9=5+117=122\). Hence, 122 is correct. Getting 124 would result from an arithmetic error. Exam tip: use \(n-1\), not \(n\), when finding the \(n\)th term.

The (10)th term is (6d) after the (4)th term, so (21+36=57). Use the difference in term numbers directly.

In this AP, the first term is \(a=200\) and the common difference is \(d=190-200=-10\). Using \(a_n=a+(n-1)d\), we get \(a_{16}=200+(16-1)(-10)=200-150=50\). Hence, 50 is correct. Choosing 60 would count only 14 differences, whereas there are 15 differences from the first term to the 16th term. Exam tip: always use \(n-1\) in the nth-term formula and keep track of a negative common difference.

The nth term of an AP is \(a_n=a+(n-1)d\). Substituting \(a=0\), \(n=11\), and \(d=11\), we get \(a_{11}=0+(11-1)\times11=110\). Choosing \(111\) would result from incorrectly using \(n\times d\), whereas the formula uses \(n-1\). Exam tip: write \(n-1\) first before substituting values in the nth-term formula.

Here (d=\frac{1}{2}), so (a_9=\frac{5}{4}+8\times\frac{1}{2}=\frac{21}{4}). Use the fractional difference carefully.

For this AP, the first term is \(a=31\) and the common difference is \(d=34-31=3\). Use \(a_n=a+(n-1)d\). Thus, \(a_{22}=31+(22-1)\times3=31+63=94\). Therefore, 94 is the correct answer. Choosing 91 means adding only 20 common differences, so it is the 21st term. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{30}=5+(30-1)(-1)=5-29=-24\). Therefore, \(-24\) is correct. The value \(-25\) results from incorrectly using \(30d\) instead of \(29d\). Exam tip: always use \((n-1)d\) in the formula for the \(n\)th term.

For this AP, the first term is \(a=18\) and the common difference is \(d=28-18=10\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_9=18+(9-1)\times10=18+80=98\). The answer 108 would result from adding 9 differences, but only 8 differences are added to reach the 9th term. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.