Introduction to APs and common difference.

In an arithmetic progression, twice the middle term equals the sum of the first and third terms. Thus, \(2(4m+2)=3m+(6m-2)\). This gives \(8m+4=9m-2\), so \(m=6\). If 5 is substituted, the two consecutive differences are not equal. Exam tip: for three AP terms, use \(2b=a+c\) directly.

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 sequence is an arithmetic progression for every value of q, so A is correct. Subtract consecutive terms: (8+3q)−8=3q, (8+6q)−(8+3q)=3q, and (8+9q)−(8+6q)=3q. Since the same algebraic difference occurs each time, the sequence is an AP, with common difference d=3q. This remains true even when q=0, when all terms become 8; a constant sequence is an AP with d=0. B is wrong because q need not equal 1. C is wrong because q=0 is only one special case, not the only case. D is wrong because equal successive differences prove that it is an AP. Memory cue: compare consecutive subtractions symbolically, not by choosing one value of q.

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.

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

The direct answer is option B: d = −4. The displayed terms are 6, 6 + d, and 6 + 2d, and they are said to become 6, 2, −2. Match corresponding terms. The first term already agrees. From the second term, 6 + d = 2. Subtracting 6 from both sides gives d = −4. Verify with the third term: 6 + 2(−4) = 6 − 8 = −2, exactly as required. Option A, 4, would make the terms 6, 10, 14, increasing rather than decreasing. Option B, −4, gives the required sequence and is correct. Option C, 8, gives 6, 14, 22. Option D, −8, gives 6, −2, −10, so the third term is wrong. A negative common difference means that the terms decrease by its positive magnitude.

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).

The common difference between consecutive terms is constant. Here, the difference between 10 and 18 is \(18-10=8\). Therefore, the term immediately before 10 is \(10-8=2\), so \(s=2\). Check: in \(-6,2,10,18\), 8 is added each time. Exam tip: first find the common difference using two known adjacent terms near the missing term.

The original (d=4); doubling makes (d=8), and subtracting the same (1) does not change (d). Multiplication changes (d), equal subtraction does not.

The fourth term is \(4+3d\). Given \(4+3d=31\), we get \(3d=27\Rightarrow d=9\). The second term is \(4+d=4+9=13\).

(18-\frac{3}{2}=\frac{33}{2}), and (-\frac{3}{2}) continues to be added. Check both (a) and (d) together.

From (2(3u+2)=(2u-3)+(5u-1)), (6u+4=7u-4), so (u=8). Use the twice-middle-term rule to find the variable.

When all terms are multiplied by (-\frac{1}{2}), (d) is also multiplied by it, so the new (d=-5). The multiplier applies directly to the common difference.