Finding the $n$th term of an AP

Correct answer: B. (43)

(a_1+a_2=a+(a+4)=18), so (a=7). Now (a_{10}=7+9\times4=43). First find the first term, then the nth term.

Correct answer: B. (30)

In an AP, the average of equally spaced terms is the middle term, so (a_7=\frac{60}{2}=30). For symmetric terms, the average method is faster.

Correct answer: C. 16

The sequence is an arithmetic progression with first term a = 64 and common difference d = 58 − 64 = −6. Since the terms decrease by 6, we need the final term that is still strictly greater than 10. Near the end, the relevant consecutive terms are 22, 16, and 10: 16 > 10, but the next term is 16 − 6 = 10, which is not greater than 10. Therefore 16 is the last permissible term, so option C is correct. Option A is excluded because it is equal to 10 rather than greater; option B is not a term of this AP; and option D is an earlier term, not the last one satisfying the condition. The strict word “greater” is essential here.

Correct answer: B. (193)

The direct answer is option B, 193. In an arithmetic progression, a is the first term and d is the fixed increase. Here a=4 and d=11-4=7. To reach the 28th term, we make 27 equal jumps, because the first term needs zero jumps. Thus a_28=a+27d=4+27(7)=4+189=193. Option A, 190, is too small and does not follow the sequence. Option B, 193, is exactly the calculated 28th term. Option C, 196, would result from adding one extra jump, effectively using 28d instead of 27d. Option D, 199, is even farther away and also uses too many increases. Another check is that the sequence is 4, 11, 18, and every step adds 7; after 27 steps the result is 193. Exam cue: the term number is not the number of differences; subtract one first.

Correct answer: A. −42

Use the nth-term formula a_n = a + (n − 1)d. The first term is a = 38, and the common difference is d = 33 − 38 = −5. For n = 17, a_17 = 38 + (17 − 1)(−5) = 38 + 16(−5) = 38 − 80 = −42. Hence option A is correct. The negative difference must be retained because the AP decreases by 5 at each step. A frequent error is to use +5, which would produce a positive result, or to multiply by 17 instead of 16. Checking by counting 16 decreases from 38 gives 38 − 80 = −42. The neighboring options do not satisfy the formula for the 17th position.

Correct answer: C. 19

The formula for the nth term of an AP is \(a_n=a+(n-1)d\). Substituting the given values gives \(121=13+(n-1)\times6\). Thus, \(108=6(n-1)\), so \(n-1=18\) and \(n=19\). Option 18 is the value of \(n-1\), not of n. Exam tip: always add 1 after finding \(n-1\).

Correct answer: A. First term 7 and common difference -3

Comparing with \(a_n=a+(n-1)d\), we get \(a=7\) and \(d=-3\). Option B misses the negative sign. Exam tip: identify the coefficient of \(n-1\) as \(d\).

Correct answer: C. 17th

The nth-term rule for an arithmetic progression is a_n = a + (n − 1)d. Here a = 21 and d = 29 − 21 = 8. Set the nth term equal to the given value 149: 149 = 21 + (n − 1)8. Subtracting 21 gives 128 = 8(n − 1), so n − 1 = 16 and n = 17. Therefore 149 is the 17th term, making option C correct. The result is easily verified: the 17th term is 21 + 16 × 8 = 21 + 128 = 149. The 16th term is 141 and the 18th term is 157, so the adjacent choices cannot be correct. The factor n − 1 is necessary because the first term already corresponds to n = 1.

Correct answer: B. 123

For an arithmetic progression, use a_n = a + (n − 1)d. The first term is a = −15, and the common difference is d = −9 − (−15) = 6. Substituting n = 24 gives a_24 = −15 + (24 − 1)6 = −15 + 23 × 6 = −15 + 138 = 123. Thus option B is correct. The negative sign belongs to the first term, while the common difference is positive because the sequence increases by 6 each time. A likely mistake is to write −15 + 24 × 6, which counts one extra gap and gives 129, or to mishandle the negative first term. Checking the sequence rule confirms that 23 jumps from the first term lead to the 24th term.

Correct answer: A. 6

For an AP, \(a_{14}-a_6=(14-6)d\). Thus, \(79-31=8d\), so \(48=8d\) and \(d=6\). If the common difference were 4, the difference over 8 positions would be only 32, not the given 48. Exam tip: when two non-consecutive terms are given, divide the difference of the terms by the difference of their indices.

Correct answer: D. (107)

The direct answer is D, 107. In an AP, the difference between two terms equals the difference in their positions multiplied by the common difference. From the given values, \(a_{13}-a_5=67-27=40\), while the position difference is \(13-5=8\). Hence \(8d=40\), so \(d=5\). The 21st term is eight positions after the 13th term, so \(a_{21}=a_{13}+(21-13)d=67+8\times5=67+40=107\). Option A, 99, is too small and does not add all eight increases. Option B, 103, misses one increase. Option C, 111, adds too many increases. Option D, 107, is correct. A helpful check is that the sequence rises by 5 each time, and moving eight places forward must increase 67 by 40.

Correct answer: A. −3

The governing formula for the nth term of an arithmetic progression is a_n=a_1+(n−1)d. Here the first term is a_1=96 and the common difference is d=87−96=−9. The negative difference confirms that each term decreases by 9. For the 12th term, eleven changes occur after the first term, so a_12=96+(12−1)(−9)=96−99=−3. Thus option A is correct. A common error is to use twelve differences, although the first term is already counted; that would produce the wrong value. Options B, C, and D also ignore the continued decrease or mishandle the negative sign. Substitution into the formula is the direct and reliable method.

Correct answer: C. 4

For an AP, the nth-term formula is \(a_n=a+(n-1)d\). Here, \(93=25+(18-1)d\), so \(93=25+17d\). Thus, \(68=17d\) and \(d=4\). If the common difference were 3, the 18th term would be \(25+17\times3=76\), not 93. Exam tip: use \(n-1\), not n, in the nth-term formula.

Correct answer: B. 117

From a known term of an AP, use \(a_n=a_m+(n-m)d\). Thus, \(a_{18}=a_4+(18-4)d=19+14\times7=19+98=117\). Therefore, 117 is correct. Choosing 119 would result from using an incorrect number of common differences. Exam tip: first find the difference between the term indices, then multiply it by \(d\).

Correct answer: D. 28.2

For an arithmetic progression, the nth term is aₙ = a + (n − 1)d. The first term is a = 1.2, and the common difference is d = 2.7 − 1.2 = 1.5. The next difference, 4.2 − 2.7, is also 1.5, confirming the AP. For n = 19, substitute into the formula: a₁₉ = 1.2 + (19 − 1)(1.5) = 1.2 + 18 × 1.5 = 1.2 + 27 = 28.2. Thus option D is correct. A common mistake is to multiply the difference by 19 rather than 18, which gives 29.7; that incorrectly counts one extra step because the first term already occupies position one. The decimal calculation and the index both support 28.2.

Correct answer: C. 138

Given \(a_n=5n+8\). To find the 26th term, substitute \(n=26\): \(a_{26}=5\times26+8=130+8=138\). Therefore, 138 is correct. The value 136 would result from incorrectly adding 6 instead of 8. Exam tip: substitute the term number carefully for \(n\) in the direct formula.

Correct answer: A. -10

The given direct formula is \(a_n=82-4n\). Substituting \(n=23\), we get \(a_{23}=82-4(23)=82-92=-10\). Therefore, the correct answer is \(-10\). A value such as \(-8\) can result from an incorrect multiplication or subtraction. Exam tip: In a direct formula, substitute the term number for \(n\), multiply first, and then subtract.

Correct answer: B. (\frac{63}{4})

Here (a=\frac{3}{4}) and (d=\frac{1}{2}) so (a_{31}=\frac{3}{4}+30\cdot\frac{1}{2}=\frac{63}{4}). Use a common denominator for fractions.

Correct answer: D. 17

For an arithmetic progression, the difference between consecutive terms is constant. Its nth term is given by \\(a_n=a_1+(n-1)d\\), where \\(a_1\\) is the first term and \\(d\\) is the common difference. Here the tenth term is 62 and the common difference is 5, so the formula can be used to work backward to the first term.

Substitute the known values: \\(62=a_1+(10-1)5\\). Since \\(10-1=9\\), the added part is \\(9\\times5=45\\). Therefore \\(62=a_1+45\\), and subtracting 45 from both sides gives \\(a_1=62-45=17\\). Thus option D is correct. The result also makes sense because moving from the first term to the tenth term requires nine equal increases of 5, whose total increase is 45.

Correct answer: C. (20)th

For an arithmetic progression, the nth term is given by a_n=a+(n-1)d, where a is the first term and d is the common difference. Here a=14 and d=23-14=9. To find which term equals 185, substitute these values: 185=14+(n-1)9. Subtracting 14 gives 171=9(n-1), and dividing by 9 gives n-1=19. Therefore n=20.

A direct check is also useful. The twentieth term is 14+19×9=14+171=185, so the number is reached exactly at the twentieth position. Thus option C, the 20th term, is correct. The 19th term would be 176 and the 21st term would be 194, so neither neighboring option can represent 185. The common difference must be used with n-1, because the first term already corresponds to n=1.

Correct answer: B. (115)

(a=-32) and (d=7) so (a_{22}=-32+21\times7=115). Be careful while adding from a negative start.

Correct answer: D. (164)

(d=\frac{104-44}{10}=6) so (a_{28}=104+10\times6=164). Add the same term difference for equal position gaps.

Correct answer: A. 4

The nth-term formula of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_{26}=16+25d\). Given \(a_{26}=116\), we get \(116=16+25d\), so \(25d=100\) and hence \(d=4\). If 3 were used, the 26th term would be \(16+25\times3=91\), not 116. Exam tip: in \(a_n\), the common difference is multiplied by \(n-1\), not by \(n\).

Correct answer: A. First term \(-3\) and common difference \(4\)

Putting \(n=1\) gives \(a_1=4(1)-7=-3\). The coefficient of \(n\) is 4, so each successive term increases by 4; hence the common difference is 4. Exam tip: always verify the first term using \(n=1\).

Correct answer: A. 145

Here, the first term is 5 and the common difference is 7. The nth term is \(a_n=5+7(n-1)=7n-2\). From \(7n-2<150\), we get \(n<\frac{152}{7}\), so the greatest integer value of \(n\) is 21. Hence, \(a_{21}=7(21)-2=145\). The numbers 146, 147, and 148 are not terms of this AP because they do not occur when starting from 5 and adding 7 each time. Exam tip: for “less than,” use a strict inequality, not equality.