Finding the $n$th term of an AP

In this AP, the first term is \(a=-3\) and the common difference is \(d=1-(-3)=4\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_7=-3+(7-1)\times4=-3+24=21\). Hence, 21 is correct. The option 17 is the sixth term and may result from incorrectly skipping the first term while counting. Exam tip: always use \((n-1)\), not \(n\), in the nth-term formula.

In this AP, the first term is \(a=9\) and the common difference is \(d=6-9=-3\). Using \(a_n=a+(n-1)d\), we get \(a_{11}=9+(11-1)(-3)=9-30=-21\). \(-18\) is the 10th term, so it is a close but incorrect option. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

The formula for the \(n\)th term of an AP is \(a_n=a+(n-1)d\). Substituting \(a=4\), \(d=7\), and \(n=5\), we get \(a_5=4+(5-1)\times7=4+28=32\). Hence, \(32\) is correct. \(35\) would result from adding \(5d\), but the fifth term requires adding only \(4d\). Exam tip: always use \((n-1)d\) for the \(n\)th term.

Here (a=1) and (d=3), so (a_{15}=1+14\times3=43). Taking one less than the term number is necessary.

For this AP, the first term is \(a=30\) and the common difference is \(d=25-30=-5\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_6=30+(6-1)(-5)=30-25=5\). Therefore, 5 is the correct answer. The value 10 can result from using the common difference incorrectly. Exam tip: in a decreasing AP, the common difference is usually negative.

The \(n\)th term of an arithmetic progression is \(a_n=a_1+(n-1)d\). Thus, \(a_{20}=8+(20-1)\times2=8+38=46\). Therefore, the correct answer is \(46\). \(48\) would result from incorrectly adding \(20d\) instead of \((20-1)d\). Exam tip: always use \((n-1)\) in the formula for the \(n\)th term.

For this AP, the first term is \(a=13\) and the common difference is \(d=17-13=4\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_9=13+(9-1)\times4=13+32=45\). Hence, 45 is correct. The value 43 would result from adding 4 only seven times, whereas the 9th term requires eight common differences. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

The first term is \(a=-10\), and the common difference is \(d=-7-(-10)=3\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{14}=-10+(14-1)\times3=-10+39=29\). Hence, \(29\) is correct. A close error such as \(26\) may result from an incorrect multiplication or from mishandling the negative first term. Exam tip: use \(n-1\), not \(n\), in the nth-term formula.

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Here, \(a=15\), \(d=-4\), and \(n=8\). Therefore, \(a_8=15+(8-1)(-4)=15-28=-13\). Hence, \((-13)\) is correct. \((-11)\) may result from using the wrong number of steps or forgetting \((n-1)\). Exam tip: when the common difference is negative, each successive term decreases.

In this AP, the first term is \(a=6\) and the common difference is \(d=13-6=7\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{10}=6+(10-1)\times7=6+63=69\). Hence, option B is correct. The value 67 would result from adding 61, but the 10th term requires adding 9 common differences to the first term. Exam tip: for term number \(n\), always use \((n-1)d\).

In this AP, the first term is \(a=100\) and the common difference is \(d=90-100=-10\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{12}=100+(12-1)(-10)=100-110=-10\). Option 0 is the 11th term, as it uses only 10 common differences from the first term. Exam tip: for the \(n\)th term, use \(n-1\), not \(n\), differences.

The nth term of an AP is \(a_n=a+(n-1)d\). Here, \(a=3\), \(d=9\), and \(n=7\). Thus, \(a_7=3+(7-1)\times9=3+54=57\). Therefore, 57 is correct. The value 54 is only \(6d\); the first term 3 has not been added to it. Exam tip: Always use \(n-1\), not \(n\), in the nth-term formula.

In this AP, the first term is \(a=11\) and the common difference is \(d=14-11=3\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{18}=11+(18-1)\times3=11+51=62\). Hence, 62 is correct. A value such as 60 can result from an incorrect term count. Exam tip: use \(n-1\), not \(n\), when finding the \(n\)th term of an AP.

The first term of this AP is 50 and the common difference is \(d=47-50=-3\). Using \(a_n=a+(n-1)d\), \(a_{13}=50+(13-1)(-3)=50-36=14\). Therefore, 14 is correct. The answer 11 can result from counting 13 differences instead of \(n-1\) differences. Exam tip: for the \(n\)th term of an AP, always use \(n-1\) common differences.

Here, the first term is \(a=4\) and the common difference is \(d=9-4=5\). The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{16}=4+(16-1)\times5=4+75=79\). Hence, \(79\) is correct. \(81\) would be obtained for the \(17\)th term, not the \(16\)th term. Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

In this AP, the first term is \(a=0\) and the common difference is \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_{21}=0+(21-1)\times5=100\). Therefore, the correct answer is 100. Note that 95 is the 20th term. Exam tip: Use \(n-1\) gaps, not \(n\), when finding the \(n\)th term.

Here (a=\frac{1}{2}) and (d=\frac{1}{2}), so (a_{10}=5). Use a common denominator while adding fractions.

For this AP, the first term is \(a=3\) and the common difference is \(d=0\). Thus, \(a_{25}=a+(25-1)d=3+24\times0=3\). Hence, the 25th term is 3. \(75\) would result only if the terms increased by 3 each time. Exam tip: In a constant AP, \(d=0\), so every term equals the first term.

The nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{17}=25+(17-1)(-2)=25+16(-2)=25-32=-7\). Hence, \(-7\) is correct. An answer such as \(-5\) may result from using the wrong term count instead of \(n-1\). Exam tip: calculate \(n-1\) first and handle the negative common difference carefully.

In this AP, the first term is \(a=8\) and the common difference is \(d=15-8=7\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{11}=8+(11-1)\times7=8+70=78\). Hence, \(78\) is correct. Getting \(76\) usually results from using an incorrect number of differences instead of \(n-1\). Exam tip: for the \(n\)th term, always add \(n-1\) common differences to the first term.