Introduction to APs and common difference.

In an arithmetic progression with three terms, the middle term equals the average of the first and third terms. Thus, \(h=\frac{4+20}{2}=12\). Therefore, \(h-4=12-4=8\). Note that \(12\) is the value of \(h\), whereas the question asks for \(h-4\). Exam tip: for three AP terms, use middle term = average of the end terms.

The original (d=4), and (3n) has (d=3), so the new (d=1). In term-number changes, subtract its common difference.

The defining condition for three consecutive AP terms is that the middle term is the arithmetic mean of the other two. Thus 2(k+2)=(k−3)+(2k+1). Simplifying, 2k+4=3k−2, so k=6. A direct check gives the terms 3, 8, and 13 when k=6; their differences are 8−3=5 and 13−8=5. Therefore option C is correct. The nearby values 5 and 7 are plausible distractors because they can result from an arithmetic rearrangement error, but neither produces equal consecutive differences.

In an arithmetic progression, the middle term is the average of the extremes, so (2(5k+1)=(3k-2)+(8k-3)). For three terms, this is the fastest exam rule.

The consecutive differences are (2d) and (3d), which cannot be equal when (d \neq 0). In exams, compare consecutive differences directly.

In an arithmetic progression, consecutive terms have the same difference. There are 4 gaps between the first and fifth terms, so the common difference is \(d=\frac{39-7}{4}=8\). Thus, \(x=7+8=15\) and \(y=23+8=31\). Therefore, \(x+y=15+31=46\). Option 42 can result from counting the gaps incorrectly. Exam tip: between \(n\) terms, there are always \(n-1\) gaps.

In an arithmetic progression, the differences between consecutive terms are equal. Therefore,
\((2p+1)-(p-4)=(4p-2)-(2p+1)\).
This gives \(p+5=2p-3\), so \(p=8\). The common difference is then \((2p+1)-(p-4)=17-4=13\). Values such as 15 do not equal the consecutive difference of the given terms. Exam tip: for three terms in an AP, set twice the middle term equal to the sum of the first and third terms.

The statement becomes false when the three displayed terms do form an arithmetic progression. Apply the middle-term condition: 2(3m−2)=(m+1)+(6m−8). This gives 6m−4=7m−7, and therefore m=3. Substituting m=3 produces the terms 4, 7, and 10, with differences 3 and 3. Hence the corrected answer is 3. None of the original options contained 3, so the options have been repaired by replacing the distractor 6 with 3. The other listed values do not make the two consecutive differences equal.

In a four-term arithmetic progression, the change from the first term to the fourth term contains three equal common differences. Let the common difference be d. Then 4+3d=31, so 3d=27 and d=9. Since r is the second term and s is the third term, their difference is s−r=d=9. Equivalently, the terms are 4, 13, 22, and 31, which confirms the result directly. Therefore option C is correct. The number 27 is the total change from the first to fourth term, not the difference between adjacent terms, so it is not an answer choice.

In an arithmetic progression, the differences between consecutive terms must be equal. Here, the first difference is \(q^2+q-q^2=q\), and the second difference is \(q^2+3q-2-(q^2+q)=2q-2\). Thus, \(q=2q-2\), which gives \(q=2\). Therefore, option 2 is correct. For instance, when \(q=1\), the differences are 1 and 0, so the terms are not in AP. Exam tip: for three terms \(a,b,c\) in AP, you may also use \(2b=a+c\).

In an AP, the difference between every pair of consecutive terms must be constant. Here, 9−4=5 and 14−9=5, but 20−14=6. So it is not an AP. Exam tip: check all consecutive differences.

In an arithmetic progression, twice the middle term equals the sum of the first and third terms. Thus, \(2(5a-1)=(2a+3)+(11a-13)\). This gives \(10a-2=13a-10\), so \(a=\frac{8}{3}\). Hence, the second term is \(5a-1=5\times\frac{8}{3}-1=\frac{37}{3}\). The value \(14\) results from the incorrect calculation \(a=3\). Exam tip: for three AP terms, use \(2b=a+c\) directly.

In an arithmetic progression, consecutive differences must be equal. Here, second term − first term is \((18-c)-18=-c\), while third term − second term is \((18-3c)-(18-c)=-2c\). Therefore, \(-c=-2c\), which gives \(c=0\). For \(c=1\) or \(c=3\), the two differences are not equal. Exam tip: For three terms to be in AP, equate their two consecutive differences.

The differences are (t, 2t, 3t), and all are equal only when (t=0). In such questions, check every consecutive difference.

From \(p+r=2q\), we get \(q-p=r-q\); from \(q+s=2r\), we get \(r-q=s-r\). Thus all consecutive differences are equal. Option B alone is insufficient. Exam tip: compare consecutive differences.

In an arithmetic progression, the difference between every pair of consecutive terms is the same. Here the four terms are 12, b, 2b−3, and 39. Because there are four terms, the change from the first term to the fourth term contains three equal common differences. This lets us determine the common difference before finding b.

The total change is 39−12=27. Therefore, the common difference is \(d=27/3=9\). The second term is one common difference after the first, so \(b=12+9=21\). Checking the next term gives \(2b−3=2(21)−3=39\), which is consistent with the fourth term. Thus option C, 21, is correct. A value such as 20 would not produce equal consecutive differences.

For the three terms to form an AP, their consecutive differences must be equal. The first difference is \((4x+1)-(2x+5)=2x-4\), and the second difference is \((7x-7)-(4x+1)=3x-8\). Equating them, \(2x-4=3x-8\), gives \(x=4\). Option 3 is a close distractor, but it does not make the two differences equal. Exam tip: For three algebraic terms in an AP, calculate the second-minus-first and third-minus-second differences before equating them.

First calculate the two consecutive terms from the given rule. A_15=4(15)−9=60−9=51, while A_14=4(14)−9=56−9=47. Hence A_15−A_14=51−47=4. More generally, substituting n+1 and n gives A_{n+1}−A_n=[4(n+1)−9]−[4n−9]=4, a constant independent of n. The difference between consecutive terms of an arithmetic progression is its common difference, so option B is correct. It is not the first term, the number of terms, or a last term.

In (20,16,12,8), (4) is subtracted each time, so (d=-4). Do not just see decreasing order; check equal differences too.

In a three-term arithmetic progression, the first and third terms sum to twice the middle term. Here (2(a+d)=0).