Introduction to APs and common difference.

From (2y=9+(2y+6)), we get (0=15), so it never forms an arithmetic progression. None of the listed values can be its common difference.

For an arithmetic progression, (2y=9+(2y+6)) is required, which gives the impossible (0=15). When an impossible equation appears, no value works.

For three consecutive terms u, v, w of an arithmetic progression, the middle term is the average of the two outer terms. Thus, 2v = u + w = 48, so v = 24. Option 22 is not correct because it would give 2v = 44. Exam tip: For any three consecutive AP terms, use middle term = (first term + third term)/2.

In an arithmetic progression, the differences between consecutive terms must be equal. Here the differences are \(k\), \(3k\), and \(5k\). For them to be equal, \(k=3k=5k\), which is possible only when \(k=0\). Then the sequence becomes \(2,2,2,2\), with common difference \(0\). For example, when \(k=1\), the differences are \(1,3,5\), so it is not an AP. Exam tip: To test an AP, compare all consecutive differences.

The common difference of an arithmetic progression is obtained by subtracting a term from the next term. Here, \((14-d)-14=-d\). Also, \((14-2d)-(14-d)=-d\), so the common difference is \(-d\). Although the terms decrease by \(d\), the difference must carry a negative sign. Exam tip: always calculate common difference as next term minus previous term.

In an arithmetic progression, the differences between consecutive terms must be equal. Here, the difference between the second and first terms is \((2n+5)-(n+2)=n+3\), while the difference between the third and second terms is \((4n+8)-(2n+5)=2n+3\). Thus, \(n+3=2n+3\), giving \(n=0\). On substituting \(n=0\), the terms are \(2,5,8\), with common difference \(3\). For example, at \(n=1\), the differences are \(4\) and \(5\), so it is not an AP. Exam tip: For three terms to form an AP, equate the two consecutive differences.

For three consecutive terms of an AP, twice the middle term equals the sum of the first and third terms. Thus, \(2(z+10)=3z+(22-z)\). On simplifying, the left side is \(2z+20\) and the right side is \(2z+22\), which gives \(20=22\), an impossibility. Hence, no value of \(z\) satisfies the condition, so the first term cannot be determined. Exam tip: For three AP terms, always check \(2b=a+c\).

This sequence has ratio (\frac{1}{3}), but differences (-54, -18, -6) are not equal. In an arithmetic progression, check difference, not ratio.

The defining condition of an arithmetic progression is that the difference between every pair of consecutive terms must be the same constant. Calculate the successive differences: 5-2=3, 10-5=5, and 17-10=7. Since 3, 5, and 7 are unequal, no single common difference exists, so the sequence is not an arithmetic progression. Option C states this exact reason and is correct. Merely increasing terms do not guarantee an AP; for example, their gaps may change, as they do here. Positive differences alone are also insufficient, so option B is false. The size of the first term has no bearing on the definition, making option D irrelevant.

In an arithmetic progression, the differences between consecutive terms must be equal. Here, the difference between the second and first terms is \((2r+2)-(r-1)=r+3\), while that between the third and second terms is \((4r+7)-(2r+2)=2r+5\). Thus, \(r+3=2r+5\), giving \(r=-2\). The terms then become \(-3,-2,-1\), so the common difference is \(d=1\). For \(r=0\), the two differences are 3 and 5, so that option is not correct. Exam tip: for three AP terms, also check whether \(2\times\) the middle term equals the sum of the first and third terms.

Use the condition for three consecutive arithmetic-progression terms: twice the middle term equals the sum of the first and third terms. Thus 2(x+6)=x+(3x−2). Expanding gives 2x+12=4x−2, so 14=2x and x=7. The third term is therefore 3x−2=3(7)−2=21−2=19. Hence option A is correct. The original options and explanation were inconsistent: the calculation actually gives x=7 and third term 19, not x=9 or 25. The corrected options now include 19 and retain distinct distractors.

In an arithmetic progression, the differences between consecutive terms are equal. Here, the difference between the second and first terms is \((x+6)-x=6\), while that between the third and second terms is \((3x-2)-(x+6)=2x-8\). Therefore, \(2x-8=6\), so \(2x=14\) and \(x=7\). If \(x=6\), the second difference becomes 4, not 6. Exam tip: For three AP terms, you can also use \(2\times\) middle term = first term + third term.

In an arithmetic progression, the differences between consecutive terms are equal. The first difference is \((s+3)-\frac{s}{2}=\frac{s}{2}+3\), and the second is \((3s-1)-(s+3)=2s-4\). Thus, \(\frac{s}{2}+3=2s-4\), which gives \(s=\frac{14}{3}\). If \(s=8\), the two differences are not equal. Exam tip: For three terms in an AP, equate the first and second differences.

For three terms in an arithmetic progression, the middle term is the average of the first and third terms. Thus, \(h=\frac{6+30}{2}=18\). Hence the common difference is \(d=18-6=12\), and also \(30-18=12\). In option A, the two consecutive differences are not equal. Exam tip: For a three-term AP, quickly find the middle term by averaging the first and third terms.

In an arithmetic progression, the differences between consecutive terms are equal. Here, the second-minus-first difference is \((3x+5)-(4x-3)=-x+8\), and the third-minus-second difference is \((x+21)-(3x+5)=-2x+16\). Equating them gives \(-x+8=-2x+16\), so \(x=8\). The terms then become \(29,29,29\), making the common difference \(0\). Option 8 is the value of \(x\), not the common difference. Exam tip: first equate consecutive differences to find the variable, then substitute it to obtain \(d\).

In an arithmetic progression, the differences between consecutive terms are equal. Here, the second-minus-first difference is \((3x+5)-(4x-3)=8-x\), and the third-minus-second difference is \((x+21)-(3x+5)=16-2x\). So, \(8-x=16-2x\), giving \(x=8\). The terms then become \(29,29,29\), so the common difference is \(0\). \(8\) is only an intermediate value obtained while solving, not the common difference. Exam tip: first equate consecutive differences to find the variable, then calculate the common difference.

The differences between consecutive terms are \(2p\), \(3p\), and \(4p\). For an arithmetic progression, all these differences must be equal. From \(2p=3p\), we get \(p=0\); then all three differences are \(0\). For \(p=1\) or \(p=2\), the differences are not equal. Exam tip: To test an AP, compare the differences of consecutive terms.

In the first option, every consecutive difference is (\frac{3}{4}). With fractions, using common denominators is the safer method.

In an arithmetic progression, the differences between consecutive terms are equal. The first difference is \((7-t)-(5-2t)=t+2\), and the second difference is \((12+t)-(7-t)=5+2t\). Thus, \(t+2=5+2t\), which gives \(t=-3\). Substituting \(-2\) does not make the two differences equal. Exam tip: for three AP terms, you can also check whether twice the middle term equals the sum of the first and third terms.

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