Finding the $n$th term of an AP

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.

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 relevant concept is that terms of an arithmetic progression change by the same common difference d. From the 5th term to the 8th term there are 8 - 5 = 3 steps. Each step increases the value by 6, so the total increase is 3 × 6 = 18. Starting with a_5 = 22, we obtain a_8 = 22 + 18 = 40. Therefore option D is correct. The general formula gives the same result: a_8 = a_5 + (8 - 5)d. Options A, B, and C correspond to using too few increments or an incorrect gap between the term numbers. There is no need to find the first term because a known term can be shifted directly using the common difference.

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 governing property is the constant difference between consecutive terms of an arithmetic progression. The 9th term is 9 - 4 = 5 positions after the 4th term. Since every position increases by d = 3, the total increase is 5 × 3 = 15. Using the known term directly, a_9 = a_4 + (9 - 4)d = 18 + 5 × 3 = 33. Hence option D is correct. This method avoids unnecessarily calculating the first term. Option A is obtained by taking only four increments, while options B and C do not result from adding five equal differences of 3 to 18. The direction is forward because the requested index 9 is greater than the given index 4.

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.

In an arithmetic progression, moving from the rth term to the nth term requires n−r equal steps, each of size d. The third term is given as 14, and the seventh term is four positions later. Therefore we do not need to find the first term; we can move forward by four common differences directly.

The number of steps is 7−3=4. Since d=5, the increase is 4×5=20. Thus the seventh term is \\(a_7=a_3+(7-3)d=14+4(5)=14+20=34\\). Hence the correct choice is C. Adding only three differences would reach the sixth term, while adding seven differences would ignore that the given value is already the third term.

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.

For an arithmetic progression, the nth term is a_n = a + (n - 1)d. Here the first term is a = -8, and the common difference is d = -3 - (-8) = 5; the later pairs confirm the same difference. Therefore a_11 = -8 + (11 - 1) × 5 = -8 + 50 = 42. Thus option B is correct. The negative first term must be retained during the calculation; it is not changed to a positive 8. Option A would result from an incorrect number of jumps or an arithmetic error, while options C and D are larger than the value produced by the ten required differences. The formula works without modification even when the initial term is negative.

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.

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.