Finding the $n$th term of an AP

Correct answer: C. 26

Each term in the sequence is 4 greater than the preceding term, so it is an arithmetic progression with first term 14 and common difference 4. The fourth term is \(a_4=a_1+3d=14+3\times4=26\). Remember that the common difference is added three times to reach the fourth term.

Correct answer: C. 22

This is an arithmetic progression with common difference \(d=7-2=5\). Using \(a_5=a_1+(5-1)d\), we get \(a_5=2+4\times5=22\). Therefore, option C is correct. Exam tip: when the consecutive difference is constant, add the common difference to obtain the next term.

Correct answer: C. 24

The terms are counted from left to right: the first term is 6, the second is 12, the third is 18, and the fourth is 24. Therefore, 24 is correct, which is option C. This is an arithmetic progression with common difference 6; in an exam, count the term position carefully.

Correct answer: D. 5

Count the terms from left to right: the first term is 9, the second is 7, and the third is 5. Each successive term decreases by 2, so the third term is 5. Exam tip: Number the terms starting from 1 when identifying a term’s position.

Correct answer: C. (18) cm

This is an arithmetic progression with first term (12) and common difference (2). The fourth term is \(a_4=a_1+(4-1)d=12+3\times2=18\) cm, so option C is correct. For exams, remember that finding the fourth term requires adding the common difference three times to the first term, not four times.

Correct answer: A. 31

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{10}=4+(10-1)\times3=4+27=31\). Therefore, 31 is correct. Choosing 30 results from an incorrect use of \(n-1\). Exam tip: for the 10th term, the common difference is added 9 times to the first term.

Correct answer: B. 51

In this AP, the first term is \(a=7\) and the common difference is \(d=11-7=4\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{12}=7+(12-1)\times4=7+44=51\). Hence, 51 is correct. The number 55 is not the 12th term; it would be the 13th term. Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

Correct answer: C. -4

In this AP, the first term is 20 and the common difference is \(d=17-20=-3\). Using \(a_n=a+(n-1)d\), \(a_9=20+(9-1)(-3)=20-24=-4\). The option -1 can result from an incorrect subtraction. Exam tip: always retain the negative sign of the common difference in a decreasing AP.

Correct answer: D. 72

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{15}=2+(15-1)\times5=2+70=72\). Hence, 72 is correct. The option 75 results from incorrectly adding \(15\times5\); only 14 common differences are added to reach the 15th term. Exam tip: always use \((n-1)\), not \(n\), in the nth-term formula.

Correct answer: A. (98)

The direct answer is option A: the 20th term is 98. In an AP, the nth term is given by an = a + (n − 1)d. Here the first term is a = 3, and the common difference is d = 8 − 3 = 5. For n = 20, a20 = 3 + (20 − 1)5 = 3 + 19 × 5 = 3 + 95 = 98. Option A is correct. Option B, 95, is only the product 19 × 5 and forgets the first term. Option C, 103, comes from adding an incorrect amount or using too many differences. Option D, 100, does not follow the AP formula and is not obtained by counting the 19 equal jumps correctly. There are 19 gaps from the first term to the 20th term, not 20. Exam cue: always use n − 1, not n, in the nth-term formula.

Correct answer: B. 43

The nth term of an arithmetic progression is given by \(a_n=a+(n-1)d\). Thus, \(a_{18}=9+(18-1)\times2=9+34=43\), so the correct answer is 43. The distractor 45 results from incorrectly using \(18d\) instead of \((18-1)d\). Exam tip: always use \(n-1\) when finding the nth term from the first term.

Correct answer: C. 20

In this AP, the first term is \(a=50\) and the common difference is \(d=45-50=-5\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_7=50+(7-1)(-5)=50-30=20\). Therefore, 20 is correct. The number 25 is the sixth term, so it is a close but incorrect option. Exam tip: always retain the negative sign of \(d\) in a decreasing AP.

Correct answer: D. 76

The formula for the nth term of an AP is \(a_n=a_1+(n-1)d\). Therefore, \(a_{11}=6+(11-1)\times7=6+70=76\). Hence, option D is correct. Exam tip: multiply the common difference by \(n-1\), not by \(n\), before adding the first term.

Correct answer: A. (34)

Here (a=-5), (d=3), so (a_{14}=-5+13\times3=34). Keep signs carefully with a negative first term.

Correct answer: B. (102)

The direct answer is option B, 102. In an arithmetic progression, the difference between consecutive terms is constant. Here, the first term is a=12 and the common difference is d=18-12=6. The nth-term rule is \(a_n=a+(n-1)d\), because the first term has zero added differences. Therefore, \(a_{16}=12+(16-1)6=12+15\times6=12+90=102\). Option A, 96, is wrong because it does not result from the correct fifteenth difference. Option B, 102, is correct because it uses the first term and 15 equal jumps of 6. Option C, 108, is too large and would require 16 jumps after the first term. Option D, 90, is too small. Remember: for the nth term, always use n-1, not n, common differences.

Correct answer: C. \(-9\)

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{13}=15+(13-1)(-2)=15-24=-9\). The option \(-7\) can result from using the term number or common difference incorrectly. Exam tip: for the \(n\)th term, always use \(n-1\) common differences, not \(n\).

Correct answer: C. 121

In this arithmetic progression, the first term is \(a=1\) and the common difference is \(d=6-1=5\). Using \(a_n=a+(n-1)d\), we get \(a_{25}=1+(25-1)\times5=1+120=121\). The distractor 126 results from the common mistake of multiplying the common difference by 25 instead of 24. Exam tip: always use \(n-1\) in the formula for the nth term.

Correct answer: D. 97

For the nth term of an AP, use aₙ = a + (n − 1)d. The first term is a = 10 and the common difference is d = 13 − 10 = 3. For n = 30, a₃₀ = 10 + (30 − 1)×3 = 10 + 29×3 = 10 + 87 = 97. Thus option D is correct. The factor is 29, not 30, because reaching the 30th term from the first term requires 29 equal steps. Option A would result from an incorrect subtraction or multiplication, option B commonly comes from using 30 differences, and option C is too small. Substitution in the formula and the step count both verify 97.

Correct answer: A. \(50\)

For this AP, the first term is \(a=100\) and the common difference is \(d=90-100=-10\). The nth-term formula is \(a_n=a+(n-1)d\). Hence, \(a_6=100+(6-1)(-10)=100-50=50\). Therefore, \(50\) is correct. \(40\) would be the 7th term. Exam tip: use \(n-1\), not \(n\), in the nth-term formula.

Correct answer: B. 105

For an arithmetic progression, the formula is \(a_n=a_1+(n-1)d\). Here, \(a_1=25\), \(d=4\), and \(n=21\). Thus, \(a_{21}=25+(21-1)\times4=25+80=105\). Therefore, the correct answer is 105. Remember to multiply the common difference by \(n-1\), not by \(n\); this is why 109 is incorrect.

Correct answer: C. (58)

Here (a=-10), (d=4), so (a_{18}=-10+17\times4=58). The formula remains the same even when terms move from negative to positive.

Correct answer: D. 89

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Thus, \(a_{10}=8+(10-1)\times 9=8+81=89\). Therefore, 89 is correct. The value 90 can result from an incorrect use of the common difference. Exam tip: for the nth term, use \((n-1)\), not \(n\).

Correct answer: A. 117

The nth-term rule for an AP is aₙ = a + (n − 1)d. From the sequence, the first term is a = 5 and the common difference is d = 12 − 5 = 7; the later differences confirm this value. For the 17th term, substitute n = 17: a₁₇ = 5 + (17 − 1)×7 = 5 + 16×7 = 5 + 112 = 117. Therefore option A is correct. The number 16 appears because there are sixteen equal jumps from the first term to the seventeenth. Option B reflects a smaller multiplication, option C adds one extra increment, and option D does not follow the AP formula. The result can also be checked by continuing the sequence with an increase of 7.

Correct answer: B. (149)

The direct answer is option B: the 22nd term is 149. The first term is a = 2. The common difference is d = 9 − 2 = 7, and the later differences also equal 7. Use an = a + (n − 1)d. For n = 22, a22 = 2 + (22 − 1)7 = 2 + 21 × 7 = 2 + 147 = 149. Therefore option B is correct. Option A, 142, is 21 × 7 without adding the first term, so it is one part of the calculation rather than the term. Option C, 156, uses an excessive amount and does not follow the formula. Option D, 147, again represents 21 × 7 and omits the first term. From the first term to the 22nd term there are 21 equal jumps. Memory cue: nth term = first term + number of jumps × common difference, and number of jumps is n − 1.

Correct answer: A. −12

Use the AP nth-term formula aₙ = a + (n − 1)d. Here the first term is a = 30, while the common difference is d = 27 − 30 = −3. Thus a₁₅ = 30 + (15 − 1)(−3) = 30 + 14(−3) = 30 − 42 = −12. Hence option A is correct. The negative difference must be retained because the sequence decreases by 3 at every step. Option C would correspond to only 13 steps or an incomplete subtraction, option B over-subtracts by 3, and option D would be obtained by ignoring the actual number of steps. Counting fourteen changes from the first to the fifteenth term confirms the result.