Introduction to APs and common difference.

The common differences are (5) and (-4), so the sum sequence has (d=1). The termwise sum of two arithmetic progressions is also an arithmetic progression.

The constant sequence has (d=0), and the second has (d=3), so the new (d=0-3=-3). Even with a constant sequence, subtraction can change (d).

In an arithmetic progression, the difference between consecutive terms is constant. Since the difference between 2 and 8 is 6, the common difference is 6. Therefore, the term immediately before 2 is 2 - 6 = -4. Checking: in -10, -4, 2, 8, every consecutive difference is 6. Choosing -6 would not give equal consecutive differences. Exam tip: Find the common difference from known consecutive terms, then add or subtract it to obtain a missing term.

The original (d=6); taking half makes (d=3), and adding the same (5) does not change (d). Multiplication or division changes (d), equal addition does not.

From (2(2q+1)=(q-3)+(4q-1)), (4q+2=5q-4), so (q=6). Watch signs while applying the twice-middle-term rule.

In the first option, (\frac{11}{5}-3=-\frac{4}{5}), and the next difference is also (-\frac{4}{5}). Use common denominators while subtracting fractions.

The added squares (1,4,9,\ldots) have differences (3,5,\ldots), which are not equal, so the new sequence will not remain arithmetic. Squares of term numbers do not create equal differences.

Given \(a+3d-a=27\), the \(a\) terms cancel, giving \(3d=27\). Hence, \(d=9\). There are three equal common-difference gaps from the first term \(a\) to the fourth term \(a+3d\), so the total difference is \(3d\). If \(d=8\), then \(3d=24\), not 27. Exam tip: The difference between the first term and the \(n\)th term of an AP is \((n-1)d\).

The total difference (26-2=24) splits into three equal gaps, so (d=8) and the second term should be (10); hence (x=10), but the third (3x=30), so there is no solution.

Both differences are (3x-3) and (3x-3), so it is an arithmetic progression for every (x). Equal algebraic differences are sufficient.

The first sequence has (d=-4) and the second has (d=2), so the sum sequence has (d=-2). Add common differences in a sum.

Taking opposites means multiplying by (-1), so (d=5) becomes (-5); adding the same (4) does not change (d). Pay attention to sign change.

From (2(2k+4)=(k+1)+(4k-2)), (4k+8=5k-1), so (k=9). Identify the middle term correctly while forming the equation.

The new terms are (1,8,21,\ldots), and the differences are (7,13,\ldots), which are not equal. Multiplying by term number does not keep (d) constant.

The total difference from (4) to (19) is (15), so (d=5), the second term should be (9) and the third (14); this gives (x=7) and (x=\frac{13}{2}). Therefore no single (x) is possible.

When all terms are multiplied by (-3), (d) is also multiplied by (-3), so the new (d=21). A negative multiplier can change the sign.

From (2(3m+1)=(2m-3)+(5m-2)), (6m+2=7m-5), so (m=7) and (d=22-11=11). Find the variable first, then the consecutive difference.

The first sequence has (d=4) and the second has (d=6), so the difference sequence has (d=4-6=-2). In termwise difference, take the difference of common differences.

The direct answer is option B: \(d=5\). The displayed AP is \(6,6+d,6+2d,6+3d\), so each next term increases by d. The second term is given as 11, hence \(6+d=11\). Subtracting 6 from both sides gives \(d=5\). Check the fourth term: \(6+3d=6+3(5)=21\), exactly as required. Option A, 4, would make the second term 10, not 11. Option B, 5, makes both given conditions true, so it is correct. Option C, 6, would make the second term 12 and the fourth 24. Option D, 7, would make them 13 and 27. The common difference is the fixed amount added between consecutive terms. Memory cue: use the simplest given term first, then verify every other condition.

The direct answer is option B: \(d=2\). For three terms to be in AP, twice the middle term equals the sum of the first and third: \(2(x+2)=x+(2x+1)\). Expanding gives \(2x+4=3x+1\), so \(x=3\). The terms then are \(3,5,7\), and the common difference is \(5-3=2\), also \(7-5=2\). Option A, 1, is wrong because the terms would not have equal differences. Option B, 2, is correct. Option C, 3, is the value of x, not d; confusing these is a common mistake. Option D, 4, is unsupported and does not match the calculated difference. First find x using the AP condition, then calculate d from consecutive terms.

(25-4=21) is split into three equal gaps, so (d=7), and (x+1=11) gives (x=10). In four consecutive terms, finding (d) from first and last terms is fast.

The difference between consecutive terms is (4). In exams, for linear (a_n), the coefficient of (n) is the common difference.

The differences are not equal, so it is not an AP. In exams, decide by checking consecutive differences, not by looking at the terms only.

Equating differences gives (p+4=3p-8), so (p=6) and (d=10). For three consecutive terms, set second minus first equal to third minus second.

For three consecutive terms of an AP, the middle term is the average of the first and third terms. Hence, \(x=\frac{7+31}{2}=19\). The common difference is \(d=19-7=12\), and this is verified by \(31-19=12\). In option A, the difference from \(7\) to \(18\) is \(11\), but from \(18\) to \(31\) it is \(13\), so it is not an AP. Exam tip: for three consecutive AP terms, use \(2\times\text{middle term}=\text{first term}+\text{third term}\).