Finding the $n$th term of an AP

For this AP, the first term is \(a=1\) and the common difference is \(d=8-1=7\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{22}=1+(22-1)\times7=1+147=148\). Therefore, 148 is correct. A value such as 146 can result from counting the terms or the common difference incorrectly. Exam tip: for the \(n\)th term, use \((n-1)d\), not \(nd\).

For this AP, the first term is \(a=35\) and the common difference is \(d=30-35=-5\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{18}=35+(18-1)(-5)=35-85=-50\). Hence, \(-50\) is correct. A value such as \(-48\) results from not applying the common difference \(-5\) the required number of times. Exam tip: use \(n-1\) gaps, not \(n\), when finding the \(n\)th term.

In an arithmetic progression, moving from the rth term to the nth term requires n-r equal steps, each of size d. Here the given term is a_2=10, the required term is a_6, and d=4. The number of steps is 6-2=4, so a_6=a_2+(6-2)d=10+4(4)=10+16=26. Therefore option C is correct. A common mistake is to add only three differences or to use six differences, which would produce distractors such as 22 or 34. The formula a_n=a_r+(n-r)d avoids that indexing error and uses the supplied second term directly.

The first term of this AP is 6 and the common difference is 0, since the difference between consecutive terms is 0. Therefore, \(a_{100}=a+(100-1)d=6+99\times0=6\). Getting 60 by multiplying 6 by 10 would be incorrect; the terms of this AP do not increase. Exam tip: In a constant AP, \(d=0\), so every \(n\)th term equals the first term.

Here (d=1), so (a_7=\frac{2}{3}+6=\frac{20}{3}). Convert the whole number into a fraction before adding.

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{11}=14+(11-1)\times3=14+30=44\). Hence, 44 is correct. The value 47 results from incorrectly using \(n\) instead of \(n-1\). Exam tip: calculate \(n-1\) first, then multiply it by \(d\).

In this AP, the first term is \(a=3\) and the common difference is \(d=8-3=5\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_9=3+(9-1)\times5=3+40=43\). Hence, 43 is correct. The value 38 can result from using an incorrect term count in place of \(n-1\). Exam tip: identify the first term and common difference before substituting \(n-1\) in the formula.

In an AP, the common difference is the difference between consecutive terms. Here d = 3, so each next term increases by 3: 7, 10, 13. Option B describes d = -3. Exam tip: always check the sign of d.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{12}=6+(12-1)\times 8=6+88=94\). Hence, \(94\) is correct. \(96\) would result from adding 12 common differences, but only 11 differences are added to reach the 12th term. Exam tip: always use \((n-1)d\) for the \(n\)th term.

The first term is \(a=-4\), and the common difference is \(d=2-(-4)=6\). Using \(a_n=a+(n-1)d\), we get \(a_{13}=-4+(13-1)\times6=-4+72=68\). Therefore, \(68\) is correct. \(66\) may result from incorrectly using \(11\) in place of \(n-1\). Exam tip: always use \((n-1)d\), not \(nd\), in the nth-term formula.

For this AP, the first term is \(a=72\) and the common difference is \(d=66-72=-6\). The \(n\)th term is \(a_n=a+(n-1)d\). Hence, \(a_8=72+(8-1)(-6)=72-42=30\). Therefore, 30 is correct. The number 36 is the sixth term, not the eighth term. Exam tip: use the common difference \(n-1\) times to find the \(n\)th term.

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Here, \(a=9\), \(d=10\), and \(n=7\), so \(a_7=9+(7-1)\times10=9+60=69\). Hence, 69 is correct. The answer 79 would result from incorrectly adding \(7d\); only six common differences are added to reach the seventh term. Exam tip: always use \((n-1)\), not \(n\), in the nth-term formula.

The first term is \(a=4\), and the common difference is \(d=12-4=8\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{15}=4+(15-1)\times8=4+112=116\). Hence, \(116\) is correct. \(120\) results from incorrectly using \(n\) instead of \((n-1)\). Exam tip: put \(n=1\) in the formula to check that it gives \(a_1=a\).

The nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{14}=80+(14-1)(-5)=80-65=15\). Hence, 15 is correct. A common error is to get 20 by using 12 instead of \(n-1\). Exam tip: Keep a negative \(d\) in brackets while multiplying.

The first term of this AP is \(a=15\), and the common difference is \(d=22-15=7\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{17}=15+(17-1)\times7=15+112=127\). Hence, \(127\) is correct. Getting \(125\) usually results from counting the number of differences incorrectly; there are \(16\) differences up to the 17th term. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

Here (a=\frac{1}{3}) and (d=\frac{1}{3}), so (a_{12}=4). Keep denominators common in fractions.

The (11)th term is (6d) after the (5)th term, so (27+24=51). Moving ahead from the given term is quick.

The first term is 90 and the common difference is \(d=82-90=-8\). Using \(a_n=a+(n-1)d\), \(a_9=90+(9-1)(-8)=90-64=26\). Hence, 26 is correct. Option 34 is the 8th term because the common difference is applied only seven times there. Exam tip: for the \(n\)th term, use \(n-1\) common differences.

For this AP, the first term is \(a=2\) and the common difference is \(d=11-2=9\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{18}=2+(18-1)\times9=2+153=155\). Hence, \(155\) is correct. \(153\) is only \(17\times9\); the first term \(2\) must also be added. Exam tip: use \(n-1\), not \(n\), in the formula for the \(n\)th term.

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{16}=-12+(16-1)\times5=-12+75=63\). Hence, 63 is correct. Using \(16d\) incorrectly gives 68; there are only 15 common differences from the first term to the 16th term. Exam tip: always use \((n-1)d\) in the nth-term formula.