Finding the $n$th term of an AP

Use aₙ = a + (n − 1)d: −1 = 63 − 4(n − 1). Thus −64 = −4(n − 1), so n − 1 = 16 and n = 17. Therefore, −1 is the 17th term.

Here, the first term is 7 and the common difference is 5. The nth term is \(a_n=7+5(n-1)\). From \(7+5(n-1)<180\), we get \(n<35.6\), so the greatest integer value of n is 35. Thus, \(a_{35}=7+5(34)=177\). The next term is 182, which is not less than 180. Although 179 is less than 180, it is not a term of this AP. Exam tip: For a “last term less than” question, solve the inequality and take the greatest possible integer value of n.

For two terms of an AP, \(a_{18}=a_4+(18-4)d\). Therefore, \(a_4=130-14\times 8=130-112=18\). Hence, 18 is correct. Option 20 is close, but adding 14 common differences to it gives \(a_{18}=132\), not 130. Exam tip: to find an earlier term from a later term, subtract the position gap multiplied by \(d\).

The station distances form an AP with first term \(a=12\) km and common difference \(d=7\) km. The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{15}=12+(15-1)\times7=12+98=110\) km. Hence, 110 km is correct. The value 117 km would result from incorrectly adding \(15\times7\); there are only 14 gaps after the first station. Exam tip: always use \((n-1)\) in the nth-term formula.

The scores form an arithmetic progression with first term \(a=20\), common difference \(d=6\), and \(n=18\). The nth term is \(a_n=a+(n-1)d\). Therefore, \(a_{18}=20+(18-1)\times6=20+102=122\). Hence, 122 points are earned at the 18th level. Getting 120 usually results from using the wrong number of intervals. Exam tip: for the nth term, add \(n-1\) common differences, not \(n\).

This is an arithmetic progression with first term \(a=32\), common difference \(d=4\), and \(n=25\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{25}=32+(25-1)\times4=32+96=128\). Hence, 128 is correct. Adding 4 twenty-five times would give 132, but the increase occurs only 24 times after the first shelf. Exam tip: always use \((n-1)\), not \(n\), in the nth-term formula.

The water amounts form a decreasing AP, with first term \(a=240\) and common difference \(d=-18\). Thus, \(a_9=a+(9-1)d=240+8(-18)=96\) litres. The value \(102\) litres would count only 7 decreases, whereas there are 8 decreases from the first hour to the ninth hour. Exam tip: In \(a_n=a+(n-1)d\), use \(n-1\), not \(n\).

This forms an arithmetic progression with first term \(a=40\), common difference \(d=3\), and \(n=30\). Using \(a_n=a+(n-1)d\), we get \(a_{30}=40+(30-1)\times3=40+87=127\). Therefore, 127 push-ups will be done on the 30th day. The option 130 results from incorrectly using \(n\) instead of \(n-1\). Exam tip: From the first term to the \(n\)th term, the common difference is added \(n-1\) times.

This is an AP with first term \(a=1500\), common difference \(d=75\), and \(n=14\). Using \(a_n=a+(n-1)d\), \(a_{14}=1500+(14-1)\times75=1500+975=₹2475\). Option ₹2550 incorrectly adds the increase 14 times; by the 14th month, it has been added only 13 times. Exam tip: use \((n-1)d\), not \(nd\), for the \(n\)th term.

The common difference is 13. The 18th and 11th terms are 7 positions apart, so a₁₈ − a₁₁ = 7 × 13 = 91.

For an AP, \(a_n=a+(n-1)d\). Thus, \(a_{22}-a_{10}=12d=130-58=72\), so \(d=6\). Now \(a_{10}=a+9d\) gives \(58=a+9\times6\), hence \(a=4\). Option 6 is the common difference, not the first term. Exam tip: When two terms are given, first use the difference in their term numbers to find \(d\).

Here, the first term is \(a=z\) and the common difference is \(d=5\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(112=z+(19-1)\times5=z+90\), so \(z=22\). If 20 were used, the 19th term would be \(20+90=110\), not 112. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

The direct answer is C, 18. The first term is 73 and the common difference is \(68-73=-5\). Use \(a_n=a+(n-1)d\): \(-12=73+(n-1)(-5)\). Subtracting 73 gives \(-85=-5(n-1)\); dividing by -5 gives \(17=n-1\), so \(n=18\). Option A, 16, would give the term \(73+15(-5)=-2\), not -12. Option B, 17, corresponds to one fewer step and gives -7. Option C, 18, gives \(73+17(-5)=73-85=-12\), so it is correct. Option D, 19, takes one extra step and gives -17. The common mistake is to use n instead of n-1 in the nth-term formula. Check the answer by substituting n=18 into the formula.

The common difference is d = (103 − 47) ÷ (15 − 8) = 8. Since the 22nd term is 7 positions after the 15th term, a₂₂ = 103 + 7 × 8 = 159.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{15}=12+(15-1)\times\frac{7}{2}=12+14\times\frac{7}{2}=12+49=61\). Hence, 61 is correct. A value such as 59 can result from using an incorrect number of common differences. Exam tip: always use \(n-1\), not \(n\), in the nth-term formula.

Given \(a_n=10-3n\), set the required term equal to \(-44\): \(-44=10-3n\). Thus, \(-54=-3n\), so \(n=18\). Therefore, \(-44\) is the 18th term of the AP. The 17th term is \(a_{17}=10-51=-41\), so it is not correct. Exam tip: To find a term number, equate the given value to \(a_n\) and solve for \(n\).

For an arithmetic progression, the change between two terms equals the difference in their indices multiplied by the common difference. From the 13th to the 21st term there are 21 − 13 = 8 equal intervals. Hence a₂₁ − a₁₃ = 113 − 65 = 48 = 8d, giving d = 6. The sixth term is seven intervals before the thirteenth term because 13 − 6 = 7. Moving backward therefore gives a₆ = a₁₃ − 7d = 65 − 7(6) = 65 − 42 = 23. Thus option C is correct. This can also be checked by reconstructing the terms: a₁₃ = a₁ + 12d, so a₁ = 65 − 72 = −7 and a₆ = −7 + 5(6) = 23. Adding instead of subtracting when moving backward, or using the wrong interval count, produces the distractors.

Here the first term is 18 and the common difference is 9. Using aₙ = a₁ + (n − 1)d, 189 = 18 + 9(n − 1), so n − 1 = 19 and n = 20.

The (r + 6)th term is six positions after the rth term, so add 6 common differences: aᵣ₊₆ = (3r − 2) + 6 × 4 = 3r + 22.

From the 5th term to the 14th term, there are 9 equal gaps. Therefore, the common difference is \(d=\frac{90-27}{14-5}=\frac{63}{9}=7\). The 20th term is 6 places after the 14th term, so \(a_{20}=90+6\times7=132\). The value 126 would result from moving only 5 places and would correspond to the 19th term. Exam tip: when moving between two AP terms, count the gaps as \(n-m\).

Direct answer: Option A, 612. An arithmetic progression, or AP, is a list in which the same number is added each time. The first multiple of 12 greater than 400 is 408, because 12 × 33 = 396 and 12 × 34 = 408. Thus the first term is a = 408 and the common difference is d = 12. The nth-term rule is \(a_n=a+(n-1)d\). For the 18th term, \(a_{18}=408+(18-1)\times12=408+17\times12=408+204=612\). Checking the pattern also gives 408 as the first term and 612 after seventeen equal jumps of 12. Option A is correct because it is exactly the 18th term. Option B, 624, is the 19th term, since it is one further jump of 12. Option C, 636, is the 20th term. Option D, 648, is the 21st term. All are multiples of 12, but they come too late in the sequence. Memory cue: for the nth AP term, count n minus 1 jumps, not n jumps.

The greatest integer n for which 13n < 600 is 46, since 13 × 46 = 598 and 13 × 47 = 611. Therefore, the greatest multiple less than 600 is 598.

The given formula is \(a_n=8n+15\). Substituting \(n=36\), we get \(a_{36}=8\times36+15=288+15=303\). Therefore, 303 is the correct answer. Getting 301 would indicate an error in multiplication or addition. Exam tip: In an AP nth-term formula, directly substitute the required term number for \(n\).

Since \(a_{n+1}-a_n=11\), the common difference is \(d=11\). There are 10 term-steps from \(a_7\) to \(a_{17}\), so \(a_{17}=a_7+(17-7)d=58+10\times11=168\). Hence, option D is correct. A value such as 166 results from an error in counting the term difference or multiplying. Exam tip: use \(a_n=a_m+(n-m)d\) when one term of an AP is known.

Let the first term be \(a_1=a\). Then \(a_2=a+d=a+7\). Hence, \(a_1+a_2=a+(a+7)=41\), so \(2a=34\) and \(a=17\). Therefore, \(a_{10}=a+9d=17+9\times7=80\). Choosing 83 would result from using the term number incorrectly instead of \(n-1\). Exam tip: always use \(a_n=a+(n-1)d\) for the \(n\)th term of an AP.