Introduction to APs and common difference.

In an arithmetic progression, consecutive differences are equal: b-a=c-b. Rearranging gives a+c=2b. Hence, a-2b+c=a+c-2b=0. A common mistake is to focus only on the difference; the middle term is always the average of the first and third terms. Exam tip: for three AP terms, use 2b=a+c directly.

In an arithmetic progression, the differences between consecutive terms are equal. Here, the first difference is \((2a+1)-(a+4)=a-3\), and the second difference is \((5a-8)-(2a+1)=3a-9\). Thus, \(a-3=3a-9\), which gives \(a=3\). Therefore, option 3 is correct. Exam tip: For three terms to be an AP, equate the first and second differences.

The differences are (x-1, 2x+3, 2x+3). Equating the first two gives (x=-4), but then all differences do not stay equal.

The consecutive differences are \((10+d)-10=d\), \((10+2d)-(10+d)=d\), and \((10+4d)-(10+2d)=2d\). In an arithmetic progression, all consecutive differences must be equal. Thus, \(d=2d\), which gives \(d=0\). If \(d=2\) or \(d=4\), the last difference becomes 4 or 8 respectively, so it is not equal to the earlier difference. Exam tip: To test an AP, compare the differences between consecutive terms.

An arithmetic progression has the same difference between every pair of consecutive terms. In option A, 6-6=0 for every step, so its common difference is 0. Option B has common difference 6, and option C has common difference -6. Option D has differences 0, 6 and 0, which are not equal. Hence the constant sequence in option A is correct.

Find the differences between consecutive terms: \((3+2m)-3=2m\), \((3+4m)-(3+2m)=2m\), and \((3+6m)-(3+4m)=2m\). Since every consecutive difference is the same, the common difference is \(2m\). \(m\) is not the increase between consecutive terms; subtract consecutive terms to obtain the difference. Exam tip: In an AP, verify using \(d=a_{n+1}-a_n\).

In an arithmetic progression, the differences between consecutive terms must be equal. Here, second term − first term = (3y+10)−(5y+2)=−2y+8. Also, third term − second term = (y+18)−(3y+10)=−2y+8. Since the two differences are equal for every (y), the correct answer is every (y). Although (y=0) and (y=4) also work, the AP is not restricted to those values. Exam tip: For three terms in AP, twice the middle term equals the sum of the first and third terms.

Subtract consecutive terms directly. The difference of the first two terms is (x+3)−(x−5)=x+3−x+5=8. The difference of the last two terms is (x+19)−(x+11)=x+19−x−11=8. The variable x cancels in both calculations, so both differences are the same constant, 8. This also confirms that the displayed sequence is an arithmetic progression with common difference 8. Therefore option A is correct. Options involving x incorrectly treat the variable part as an uncancelled contribution, while 6 and 10 do not match either subtraction.

In an arithmetic progression, the differences between consecutive terms must be equal. Here, the first difference is \((2k+3)-(k+7)=k-4\), and the second difference is \((4k-5)-(2k+3)=2k-8\). Thus, \(k-4=2k-8\), which gives \(k=4\). Therefore, option C is correct. For instance, at \(k=3\), the two differences are not equal. Exam tip: For three terms to form an AP, equate the first and second consecutive differences.

In an arithmetic progression, the middle term is the average of the first and third terms. Therefore, \(2(2x+1)=13+(4x-5)\) must hold. Simplifying gives \(4x+2=4x+8\), or \(2=8\), which is impossible. Hence, for no value of \(x\) do these three terms form an AP, so no common difference exists. Exam tip: for three terms \(a,b,c\), check an AP quickly using \(2b=a+c\).

For three terms to be in an arithmetic progression, twice the middle term must equal the sum of the first and third terms: 2(2x+1)=13+(4x-5). Simplifying gives 4x+2=4x+8, hence 2=8, which is impossible for every x. Therefore no value of x makes the sequence an AP, so option D is correct.

In an arithmetic progression, the differences between consecutive terms are equal. Here, the difference between the second and first terms is \((6p-3)-(4p+1)=2p-4\), while that between the third and second terms is \((9p-10)-(6p-3)=3p-7\). Thus, \(2p-4=3p-7\), giving \(p=3\). Substituting this value, \(d=2p-4=2\). For option C, putting \(p=4\) does not make the two differences equal. Exam tip: In AP questions, equate the two consecutive differences first.

In (5,9,13,20), the first two differences are (4,4), but the third difference is (7). It is necessary to check all consecutive differences.

The consecutive differences are (r,2r,3r), which are equal only when (r=0). In such questions first write all consecutive differences.

In three consecutive terms, twice the middle term equals the sum of the other two, so (2(2x+5)=(3x-2)+(x+16)) gives (x=2). The middle-term rule is a fast exam method.

Using (2(3p+7)=(4p+1)+(p+19)) gives (p=6), so the terms become (25,25,25) and (d=0). First find the variable and then check (d).

The total difference (29-5=24) is split into three equal gaps, so (d=8), (x=13), and (y=21). For two missing terms, divide the total difference into equal gaps.

The original (d=4); multiplying by (3) makes (d=12), and subtracting the same (2) does not change (d). Equal addition or subtraction does not change (d).

The original (d=-3), and the difference of (2n) is (2), so the new (d=-3+2=-1). For term-number changes, add or subtract its difference.

For three terms in an arithmetic progression, the middle term is the average of the first and third. Therefore, 2(a+4)=(2a−5)+(4a−7). Simplifying the left side gives 2a+8, while the right side gives 6a−12. Thus 2a+8=6a−12, so 20=4a and a=5. Substitution produces the terms 5, 9, and 13, whose consecutive differences are both 4. Hence option D is correct. The other values fail the equality of the two consecutive differences and are only distractors.