Arithmetic Progression

The first four terms are 12, 20, 28, and 36. Therefore, their sum is \(12+20+28+36=96\). Option 92 is the sum of only the first three terms, \(12+20+28\), so it is incorrect. Exam tip: When an arithmetic progression has only a few terms, adding them directly is often quickest.

The nth-term formula of an arithmetic progression is \(a_n=a+(n-1)d\). Thus, \(a_6=34+(6-1)(-7)=34-35=-1\). Therefore, \(-1\) is correct. The value \(6\) results from incorrectly using 4 instead of \(n-1=5\) for the sixth term. Exam tip: when \(d\) is negative, check the sign after multiplication carefully.

Direct answer: Option C, 6. In an arithmetic progression, the difference between consecutive terms is constant. Use the formula for the nth term: \\(a_n=a+(n-1)d\\). We are not given the first term, but subtracting two terms removes it: \\(a_8-a_4=(8-4)d\\). Substitution gives \\(49-25=4d\\), so \\(24=4d\\), and therefore \\(d=6\\). There are four gaps from the fourth term to the eighth term: 4th→5th, 5th→6th, 6th→7th, and 7th→8th. Option A, 4, would give only a total increase of 16, not 24. Option B, 5, would give an increase of 20. Option C, 6, gives four increases of 6, totalling 24, so it is correct. Option D, 8, would give an increase of 32. Common mistake: count intervals, not term labels; from term 4 to term 8 there are 4 intervals.

Given \(a_n=8n+5\), \(a_2=8\times2+5=21\) and \(a_5=8\times5+5=45\). Therefore, \(a_2+a_5=21+45=66\). A value such as \(62\) can result from evaluating one of the terms incorrectly. Exam tip: substitute the value of \(n\) in each term separately before adding them.

The governing concept is the relationship between a specified term, the first term, and the common difference in an arithmetic progression: aₙ = a + (n − 1)d. For the sixth term, a₆ = a + 5d. Substituting a₆ = 41 and d = 5 gives 41 = a + 5 × 5 = a + 25. Therefore, a = 41 − 25 = 16, so option B is correct. The result can be checked by constructing the progression backward or forward: starting from 16 and repeatedly adding 5 gives 16, 21, 26, 31, 36, 41. Thus 41 is indeed the sixth term. Options A, C, and D come from subtracting an incorrect multiple of 5 or confusing the first term with an intermediate term.

The common difference of an arithmetic progression is the difference between two consecutive terms. Here, both the second and first terms are 12, so \(d=12-12=0\). Therefore, the correct answer is 0. The number 12 is the value of each term, not the common difference. Exam tip: use \(d=a_2-a_1\) to find the common difference.

The first term is (22) and (d=7), so (a_n=22+(n-1)7=7n+15). Remember (n-1) while forming the general term.

From (a_5) to (a_{10}), there are (5) gaps, so (a_{10}=33+5(4)=53). Moving forward from a nearby term is a quick method.

The first term is (20), and subtracting (5) each time gives (20,15,10,5). Form terms directly from the given (a) and (d).

In an arithmetic progression, (2(2x)=(x+1)+(3x-5)), which gives (x=6). The middle term is the average of the two surrounding terms.

The direct answer is option B, the 9th term. For an AP, use the nth-term formula \\(a_n=a+(n-1)d\\). Here the first term is 38 and the common difference is 33−38=−5. Thus \\(a_n=38+(n-1)(−5)\\). Check the terms step by step: a1=38, a2=33, a3=28, a4=23, a5=18, a6=13, a7=8, a8=3, and a9=−2. The first negative term is therefore the ninth term. Option A, 8th, is 3, which is positive, not negative. Option B, 9th, is −2 and is correct. Option C, 10th, is −7, also negative but not the first negative term. Option D, 11th, is −12 and is even later. The word “first” requires checking the preceding term as well. Memory cue: find the last non-negative term, then move one step further.

For the fourth term, substitute n=4 in the given rule: \(a_4=17-3(4)=17-12=5\). Therefore, the correct answer is 5. The value 8 is obtained when \(n=3\), so it is the third term, not the fourth. Exam tip: In an \(a_n\) question, substitute the required term number directly for n.

The average of the first three terms is (\frac{10+18+26}{3}=18). In an arithmetic progression, the average of three consecutive terms is the middle term.

This is an AP with \(a=18\), \(d=3\), and \(n=12\). Thus, \(a_{12}=18+(12-1)\times3=51\). Getting 54 means adding the difference 12 times instead of 11. Exam tip: always use \(n-1\) gaps.

The first term is \(a=5\), and the common difference is \(d=14-5=9\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_{14}=5+(14-1)\times9=5+117=122\). Hence, option D is correct. Option C, \(119\), results from using too small an increase instead of the required \(13d\) for the 14th term. Exam tip: for the \(n\)th term, remember that there are \(n-1\) common differences after the first term.

In (8,16,24,32,\ldots), (8) is added each time. To find the common difference, check the difference of two consecutive terms.

Direct answer: Option B, 160. An arithmetic progression changes by the same amount each time. Here the terms decrease by 6: \\(38-44=-6\\), \\(32-38=-6\\), and \\(26-32=-6\\). Therefore the fifth term is \\(26-6=20\\). The first five terms are 44, 38, 32, 26, and 20. Add them step by step: \\(44+38=82\\), \\(82+32=114\\), \\(114+26=140\\), and \\(140+20=160\\). Equivalently, use \\(S_n=\\frac n2[2a+(n-1)d]\\): \\(S_5=\\frac52[2(44)+4(-6)]=\\frac52(64)=160\\). Option A, 150, is too small and results from an addition or term error. Option B, 160, matches the correct sum. Option C, 170, overstates the sum, often because the negative difference is ignored. Option D, 180, is also too large and does not match the five actual terms. Remember to find the fifth term before summing and keep the negative common difference.

From (a_7) to (a_4), move back (3) gaps, so (a_4=58-3(8)=34). When moving backward, subtract the common difference.

Here, the first term is 35 and the common difference is \(-7\). Therefore, \(a_n=35+(n-1)(-7)\). Putting \(a_n=0\), we get \(35-7(n-1)=0\), so \(n=6\). Hence, 0 is the sixth term. The fifth term is 7, so it is not correct. Exam tip: To find a term’s position, substitute its value in \(a_n=a+(n-1)d\).

This is an arithmetic progression with first term 10 and common difference 2. The 12th term is \(a+(n-1)d=10+11\times2=32\) cm. Choosing 34 cm gives the 13th term instead. Exam tip: use \(n-1\), not \(n\), in the formula.

There are (6) gaps between (a_{12}) and (a_6), and (d=8), so the difference is (48). Term difference depends on the difference of positions.

The rule is \(a_n=16-2n\). Substituting \(n=1,2,3,4\) gives \(a_1=14\), \(a_2=12\), \(a_3=10\), and \(a_4=8\), respectively. Hence, the correct sequence is \(14,12,10,8\). Option B incorrectly treats \(16\) as the first term; however, at \(n=1\), the first term is \(14\). Exam tip: When a sequence is defined using \(a_n\), usually begin by substituting \(n=1\).

From (a_3) to (a_9), the increase over (6) gaps is (36), so (d=6), hence (a_6=14+3(6)=32). Find the common difference first.

Direct answer: Option B, 30 and 37. In an arithmetic progression, the common difference is found by subtracting consecutive terms. Here \\(16-9=7\\) and \\(23-16=7\\), so \\(d=7\\). To find the next term, add 7 to 23: \\(23+7=30\\). To find the term after that, add 7 again: \\(30+7=37\\). Option A, 28 and 33, uses differences of 5 and does not continue the given pattern. Option B, 30 and 37, keeps the difference equal to 7, so it is correct. Option C, 31 and 39, increases by 8 and then 8, not by 7. Option D, 32 and 41, increases by 9 and then 9, also not by 7. The order matters: the first requested next term must come immediately after 23. Exam cue: identify the common difference first, then repeatedly add it; do not guess from the answer choices.

There are (5) gaps between (a_7) and (a_2), so the difference is (5\times13=65). For term difference, look at the difference of positions.