Finding the $n$th term of an AP

Correct answer: A. 6

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.

Correct answer: C. (\frac{20}{3})

The direct answer is option C: the 7th term is \(\frac{20}{3}\). The terms are \(\frac23,\frac53,\frac83,\frac{11}{3},\ldots\). Subtract consecutive terms: \(\frac53-\frac23=1\), \(\frac83-\frac53=1\), so the common difference is \(d=1\). The first term is \(a=\frac23\). Using \(a_n=a+(n-1)d\), we get \(a_7=\frac23+(7-1)(1)=\frac23+6\). Write 6 with denominator 3: \(6=\frac{18}{3}\). Therefore \(a_7=\frac23+\frac{18}{3}=\frac{20}{3}\). Option A, \(\frac{18}{3}=6\), misses the first term’s \(\frac23\). Option B, \(\frac{19}{3}\), is one third too small. Option C matches the calculation. Option D, \(\frac{21}{3}=7\), is one third too large. Another check is that the numerators increase by 3: 2, 5, 8, 11, 14, 17, 20; hence the seventh term is \(20/3\). Memory cue: count six jumps from the first term to reach the seventh term.

Correct answer: B. 44

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 \(d\).

Correct answer: C. 43

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.

Correct answer: A. Each successive term is 3 greater than the preceding term.

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.

Correct answer: C. \(94\)

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.

Correct answer: C. \(68\)

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.

Correct answer: C. 30

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.

Correct answer: B. 69

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.

Correct answer: C. \(116\)

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 \(a_1=a\).

Correct answer: B. 15

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.

Correct answer: B. \(127\)

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.

Correct answer: D. 4

The governing concept is the nth-term formula for an arithmetic progression: aₙ = a + (n − 1)d. The first term is a = 1/3. The common difference is d = 2/3 − 1/3 = 1/3; the later terms confirm the same difference because 1 − 2/3 = 1/3. For n = 12, a₁₂ = 1/3 + (12 − 1)(1/3) = 1/3 + 11/3 = 12/3 = 4. Thus option D is correct. The multiplier is 11 rather than 12 because the first term is already present, and only eleven equal steps are needed to reach the twelfth term. The other options arise from counting the steps incorrectly or stopping at an earlier term.

Correct answer: C. 51

The governing AP principle is that moving from a known term aᵣ to another term aₙ changes the value by (n − r)d. Here the fifth term is 27 and the common difference is 4. From the fifth term to the eleventh term there are 11 − 5 = 6 equal intervals, not five or seven. Therefore a₁₁ = a₅ + (11 − 5)d = 27 + 6 × 4 = 27 + 24 = 51. Hence option C is correct. Finding the first term is unnecessary because a middle term has already been supplied. Option B would add only five differences, option D would add seven, and option A reflects a different counting or arithmetic error. Counting the intervals between the term numbers prevents the common off-by-one mistake.

Correct answer: B. 26

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.

Correct answer: C. \(155\)

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.

Correct answer: B. 63

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.

Correct answer: D. 79

In this AP, the first term is \(a=25\) and the common difference is \(d=31-25=6\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{10}=25+(10-1)\times6=25+54=79\). Therefore, 79 is correct. The value 73 would result from adding only 8 common differences, but the 10th term is reached after 9 differences. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

Correct answer: B. \(-2\)

For this AP, the first term is \(a=55\) and the common difference is \(d=52-55=-3\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Thus, \(a_{20}=55+(20-1)(-3)=55-57=-2\). Hence, the correct answer is \(-2\). An answer such as \(-4\) may result from incorrectly using \(20d\); the correct multiplier is \(n-1=19\). Exam tip: in the nth-term formula of an AP, the common difference is added \((n-1)\) times.

Correct answer: A. \(a+14d\)

The nth term of an AP is \(a_n=a+(n-1)d\). Substituting \(n=15\) gives \(a_{15}=a+14d\). The expression \(a+15d\) adds one extra common difference. Exam tip: always check the \(n-1\) factor.

Correct answer: A. 28

In this AP, the first term is \(a=100\) and the common difference is \(d=97-100=-3\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{25}=100+(25-1)(-3)=100-72=28\). Hence, 28 is correct. Getting 30 may result from using an incorrect number of common differences instead of 24. Exam tip: for the \(n\)th term, always use \(n-1\) differences.

Correct answer: B. 51

In an AP, there are 5 steps from the third term to the eighth term. Thus, \(a_8=a_3+(8-3)d=16+5\times7=51\). Therefore, the correct answer is 51. Choosing 49 would mean that the common difference has not been added the correct number of times. Exam tip: subtract the term numbers to find the number of common-difference steps between two terms.

Correct answer: B. 123

The governing concept is the nth-term formula of an arithmetic progression. The first term is a = 13, and the common difference is d = 23 − 13 = 10; the sequence increases by 10 each time. Thus a_12 = a + (12 − 1)d = 13 + 11 × 10 = 13 + 110 = 123. Therefore option B is correct. The multiplier is 11 because eleven equal gaps separate the first and twelfth terms. Option A could arise from subtracting or miscounting one step, while options C and D use twelve or more increments rather than the required eleven. The calculation also confirms the pattern: the terms are 10k + 3 for k = 1, 2, …, so the twelfth is 120 + 3 = 123.

Correct answer: C. 50

The first term is \(a=-20\), and the common difference is \(d=-15-(-20)=5\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{15}=-20+(15-1)\times5=-20+70=50\). Hence, 50 is correct. The value 45 may result from incorrectly using 13 instead of \(n-1\). Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

Correct answer: A. \(-17\)

The nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_6=18+(6-1)(-7)=18-35=-17\). Hence, \(-17\) is correct. An option such as \(-15\) may result from an error in multiplication or subtraction. Exam tip: When \(d\) is negative, keep it in brackets while substituting in the formula.