Introduction to APs and common difference.

The original (d=4), and (n) has difference (1), so the new (d=5). When adding term number, its difference is also added.

From the second to the sixth term there are (4) gaps, so (4d=24) and (d=6). Convert term distance into number of gaps.

From the first to the fifth term there are (4) gaps, so (4d=-20) and (d=-5). In a decreasing progression, (d) is negative.

The original (d=5), and (2n) has difference (2), so the new (d=5-2=3). In term-number subtraction, subtract its difference.

The consecutive differences are (3q,3q,3q). With equal algebraic differences, it is an arithmetic progression for every (q).

The original sequence is an AP with common difference d = 10 − 13 = −3; the next differences confirm this value. Multiplying every term by a constant multiplies the common difference by the same constant. Hence the new difference is d′ = (−2)(−3) = 6. For example, the transformed terms begin −26, −20, −14, −8, and each successive difference is 6. The negative multiplier reverses the order direction and changes the sign of the original difference, so the result is positive. Therefore option B is correct. Option C is the unchanged original difference, while the other options use an incorrect sign or magnitude.

From (2(2x+4)=x+4x), (4x+8=5x), so (x=8). Then (d=20-8=12), so none of the listed values is correct.

\(-\frac{7}{6}-\left(-\frac{5}{3}\right)=\frac{1}{2}\), and the next difference is also \(\frac{1}{2}\). Subtract negative fractions carefully.

For three terms to be in an AP, twice the middle term must equal the sum of the first and third terms. Thus, \(2(2m+3)=(m-1)+(4m-1)\). This gives \(4m+6=5m-2\), so \(m=8\). A nearby option such as \(m=6\) does not satisfy this equality. Exam tip: for three AP terms, directly use \(2b=a+c\).

The original (d=-5), so after division by (5), the new (d=-1). Dividing all terms by the same number divides (d) by that number too.

For three consecutive terms in an arithmetic progression, the middle term is the average of the first and third terms. Thus b = (a + c)/2 = (6 + 30)/2 = 36/2 = 18. The required quantity is not b itself but b − a, so b − a = 18 − 6 = 12. Equivalently, the total difference from 6 to 30 is 24, and it consists of two equal AP gaps; each gap is 24/2 = 12. Therefore option B is correct. Option C is the middle term b, while option D is the full end-to-end difference and option A does not follow from the AP condition.

The first three options have (d=-7), but (60,52,44,36,\ldots) has (d=-8). Check both value and sign in options.

From (6+d=2), (d=-4), and (6+2d=-2) confirms it. Match the general form with the given terms.

The first sequence has (d=4) and the second has (d=7), so the sum sequence has (d=11). In termwise addition, common differences add.

The first sequence has (d=-4) and the second has (d=3), so the difference sequence has (d=-4-3=-7). In termwise subtraction, subtract the differences too.

Both differences are (6) and (6), so it is an arithmetic progression for every (x). Equal algebraic differences are sufficient.

The original (d=-3), and term numbers have (d=1), so the new (d=-3-1=-4). When subtracting term number, its difference is also subtracted.

The new terms are (a+3,a+d+6,a+2d+9), and both differences are (d+3). The difference of the added numbers is added to (d).

The first sequence has (d=5) and the second has (d=6), so the difference sequence has (d=5-6=-1). In termwise difference, take the difference of common differences.

The constant sequence has (d=0), and the second has (d=3), so the sum has (d=3). Adding a constant sequence does not change (d).