Finding the $n$th term of an AP

Correct answer: C. \(155\)

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.

Correct answer: D. \(12.0\)

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.

Correct answer: C. 73

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.

Correct answer: A. \(-4\)

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.

Correct answer: C. 122

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.

Correct answer: C. 57

The governing concept is equal-step movement in an arithmetic progression. The tenth term is 10 − 4 = 6 positions after the fourth term. Since every step increases the value by the common difference d = 6, the required calculation is a₁₀ = a₄ + (10 − 4)d = 21 + 6 × 6 = 21 + 36 = 57. Therefore option C is correct. There is no need to determine the first term; the supplied fourth term is a convenient starting point. Options A, B, and D can result from adding five or seven differences, or from an arithmetic mistake. The standard formula gives the same result after finding a₁, but direct movement between the known and required terms is shorter and makes the interval count clear.

Correct answer: B. 50

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.

Correct answer: B. 110

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.

Correct answer: C. (\frac{21}{4})

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

Correct answer: D. 94

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.

Correct answer: C. \(-24\)

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.

Correct answer: C. 98

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.

Correct answer: B. 57

The governing concept is finding a term of an arithmetic progression from another known term. Instead of restarting from the first term, use aₙ = aᵣ + (n − r)d. Here the known term is a₂ = 12, the required term is a₇, and d = 9. The number of equal steps from the second position to the seventh position is 7 − 2 = 5, not 7. Therefore a₇ = 12 + (7 − 2)×9 = 12 + 5×9 = 12 + 45 = 57. Hence option B is correct. The distractors can result from adding the common difference the wrong number of times or from a simple arithmetic error. Counting gaps between positions prevents that mistake.

Correct answer: B. \(-12\)

For this AP, the first term is \(a=48\) and the common difference is \(d=42-48=-6\). The \(n\)th term is \(a_n=a+(n-1)d\). Hence, \(a_{11}=48+(11-1)(-6)=48-60=-12\). Therefore, \(-12\) is correct. A value such as \(-14\) may result from miscounting terms or using \(n\) instead of \((n-1)\). Exam tip: always use \((n-1)\) in the formula for the \(n\)th term.

Correct answer: B. \(25\)

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{10}=2.5+(10-1)\times2.5=2.5+22.5=25\). Hence, the correct answer is \(25\). The value \(27.5\) would result from incorrectly using \(10d\) instead of \((10-1)d\). Exam tip: Always use \((n-1)\) in the formula for the \(n\)th term.

Correct answer: B. 187

In this AP, the first term is \(a=7\) and the common difference is \(d=16-7=9\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{21}=7+(21-1)\times9=7+180=187\). Hence, 187 is correct. Getting 185 usually results from using an incorrect number of common differences. Exam tip: from the first term to the \(n\)th term, add the common difference \(n-1\) times.

Correct answer: B. 36

The governing concept is the nth-term relation for an arithmetic progression, including a negative common difference: aₙ = aᵣ + (n − r)d. The twelfth term is 12 − 8 = 4 steps after the eighth term. Hence a₁₂ = 44 + 4(−2) = 44 − 8 = 36, so option B is correct. The negative sign means the sequence decreases by 2 at every step; the relevant terms are 44, 42, 40, 38, and 36. Option A would subtract five differences, option C would subtract only three, and option D would subtract only two. Keeping the negative sign throughout the substitution is important, because changing it to positive would incorrectly make the sequence increase rather than decrease.

Correct answer: B. 127

For this AP, the first term is \(a=12\) and the common difference is \(d=17-12=5\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{24}=12+(24-1)\times5=12+115=127\). Therefore, 127 is correct. A value such as 125 results from an incorrect count of the common differences. Exam tip: always use \((n-1)d\), not \(nd\), in the nth-term formula.

Correct answer: C. \(87\)

The first term is \(a=-15\) and the common difference is \(d=-9-(-15)=6\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{18}=-15+(18-1)\times6=-15+102=87\). Hence, \(87\) is correct. A close error such as \(85\) can result from incorrectly counting the number of common differences. Exam tip: always use \(n-1\) common differences to find the \(n\)th term of an AP.

Correct answer: B. 12

The nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{13}=120+(13-1)(-9)=120-108=12\). Therefore, 12 is correct. Getting 21 is a common error caused by using the negative common difference \(-9\) with the wrong sign. Exam tip: always use \(n-1\) and keep a negative \(d\) in brackets while calculating.

Correct answer: B. 171

The first term of the AP is 6 and the common difference is \(17-6=11\). Using \(a_n=a+(n-1)d\), \(a_{16}=6+(16-1)\times11=6+165=171\). Hence, 171 is correct. A value such as 169 would result from using an incorrect common difference or term count. Exam tip: for the \(n\)th term, add \(n-1\) common differences, not \(n\).

Correct answer: B. \(-19\)

The first term is \(a=33\), and the common difference is \(d=29-33=-4\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{14}=33+(14-1)(-4)=33-52=-19\). Hence, \(-19\) is correct. The value \(-17\) results from adding \(-4\) only 12 times, which gives the 13th term. Exam tip: Always use \(n-1\), not \(n\), in the nth-term formula.

Correct answer: C. 74

The nth term of an AP is \(a_n=a_1+(n-1)d\). Thus, \(a_5=22+(5-1)\times13=22+52=74\). Therefore, 74 is correct. The value 76 results from incorrectly adding \(5d\); the common difference is added only four times to reach the fifth term. Exam tip: always use \(n-1\), not \(n\), in the formula.

Correct answer: B. 171

For this AP, the first term is \(a=9\) and the common difference is \(d=15-9=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{28}=9+(28-1)\times6=9+162=171\). Hence, 171 is correct. Getting 169 usually indicates an arithmetic error while adding \(27\times6\). Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

Correct answer: C. (88)

In an arithmetic progression, moving from the ninth term to the fifteenth term means moving forward by six positions: from term 9 to terms 10, 11, 12, 13, 14, and 15. Each forward position increases the value by the common difference 3. Thus, the total increase is six times 3, and this increase is added to the known ninth term. This avoids finding the first term unnecessarily.

The calculation is \\(a_{15}=a_9+(15-9)d\\). Substituting the given values gives \\(a_{15}=70+6\\times3=70+18=88\\). Therefore, option C is correct. The number of steps is 6, not 15, because the known term is already the ninth term; only the distance between the term numbers matters.