Introduction to APs and common difference.

Each next term is (4) less than the previous one, so (d=-4). A decreasing arithmetic progression has a negative 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.

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.

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 \(2b=a+c\).

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 terms. So (2(x+8)=(2x+1)+(3x-4)) gives (x=3).

(2(3a+6)=(5a-2)+(a+18)) gives (a=2). Then the terms (8,12,20) do not have equal differences, so no arithmetic progression is formed.

The direct answer is option A: only
\(p=0\). An arithmetic progression has the same difference between every pair of consecutive terms. Here the first differences are
\((4+p)-4=p\), then
\((4+3p)-(4+p)=2p\), and then
\((4+6p)-(4+3p)=3p\). For an AP, we need
\(p=2p=3p\). The first equality gives
\(p=0\), and this value also makes every difference zero, so the sequence becomes
\(4,4,4,4,\ldots\). Option A is correct. Option B,
\(p=1\), gives differences
\(1,2,3\), which are unequal. Option C is wrong because an arbitrary p does not make these differences equal. Option D is wrong because p=0 does work. The useful method is always to subtract consecutive terms before choosing an answer.

Consecutive terms of an arithmetic progression have the same common difference. Here, the difference between 25 and 19 is 6, so the term immediately before 19 is 19 - 6 = 13. Therefore, u = 13. Option 12 is not correct because the difference from 12 to 19 would be 7, not the common difference 6. Exam tip: Find the difference between known consecutive terms and check it in both forward and backward directions.

(\frac{5}{8}-\frac{3}{8}=\frac{2}{8}=\frac{1}{4}). When denominators are equal, subtract the numerators.

Adding the same number to all terms does not change the common difference. The original (d=20-24=-4), so the new (d) remains (-4).

The direct answer is option C: the common difference becomes negative. In the original sequence, moving forward gives
\(8-3=5\),
\(13-8=5\), and
\(18-13=5\), so the original common difference is
\(d=5\). Reversing the order gives
\(18,13,8,3,\ldots\). Now consecutive differences are
\(13-18=-5\),
\(8-13=-5\), and
\(3-8=-5\). Thus the new common difference is
\(-5\), which is negative. Option A is wrong because the direction has reversed, so +5 does not describe the new order. Option B is wrong because the terms are not equal. Option C is correct. Option D is wrong because a constant difference still exists. Memory cue: reversing an AP changes
\(d\) to
\(-d\).

Both consecutive differences are (6) and (6), so it is an arithmetic progression for every (x). Simplify differences before deciding the variable.

The common difference of the original sequence is \(5-9=-4\). Taking the opposite of every term gives \((-9,-5,-1,3,\ldots)\). The difference between consecutive terms is \(-5-(-9)=4\), so the new common difference is \(4\). The value \(-4\) is the common difference of the original sequence, not the new one. Exam tip: when every term is multiplied by \(-1\), the sign of the common difference also changes.

In the first option, every consecutive difference is (\frac{3}{5}). In fraction options, check all differences separately.

(20-2=18) is split into three gaps, so the second term should be (8) and the third (14). This gives (x=5) and (x=\frac{13}{3}), so no single value is possible.

The common difference is found by subtracting a term from the next term. Here, \(1.05-0.60=0.45\) and \(1.50-1.05=0.45\), so the common difference is \(0.45\). Choosing \(0.40\) would not give the correct difference between consecutive terms. Exam tip: Align decimal points before subtracting decimal numbers.

In an arithmetic progression, the common difference is the difference between consecutive terms. Here, \((r-3)-(r-9)=r-3-r+9=6\). Checking the next pair, \((r+3)-(r-3)=6\) as well. Therefore, \(d=6\). Option \(r\) is incorrect because the common difference must not depend on the variable. Exam tip: Find \(d\) by subtracting the first term from the second term.