Introduction to APs and common difference.

The governing property is that in a three-term arithmetic progression, the middle term is the arithmetic mean of the first and third terms. Therefore, b=(a+c)/2=(-6+18)/2=12/2=6. The requested expression is not b itself; it is b-a. Substituting the values gives b-a=6-(-6)=6+6=12. Hence option C is correct. Option A is the value of b, not b-a. Option B and option D do not satisfy the required subtraction and may result from using the endpoints incorrectly or confusing the common difference with the requested expression. The common difference is also 12, which agrees with the answer here because the first three terms are -6, 6, 18.

The first three options have (d=-9), but (70,62,54,46,\ldots) has (d=-8). Check both value and sign in the options.

The expression (5, 5+d, 5+2d) represents three consecutive terms of an arithmetic progression whose first term is 5 and common difference is d. Match the second general term with the second given term: 5+d=-1. Subtracting 5 from both sides gives d=-1-5=-6. We can verify the result using the third term: 5+2d=5+2(-6)=5-12=-7, exactly as required. Therefore option B is correct. Option A has the wrong sign, usually caused by reversing the subtraction. Option C is the magnitude of the change from 5 to -7, but it ignores the fact that two equal steps occur. Option D incorrectly doubles the 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.