Introduction to APs and common difference.

The original (d=-5), so after dividing by (4), the new (d=-\frac{5}{4}). Dividing all terms by the same number divides (d) by that number.

In an arithmetic progression, the difference between consecutive terms is constant. Therefore, for three terms, the middle term b is the average of a and c: \(b=\frac{a+c}{2}=\frac{4+22}{2}=13\). Hence, \(b-a=13-4=9\). Option 11 can result from using an incorrect difference instead of halving \(c-a=22-4\). Exam tip: For three consecutive AP terms, the middle term is always the average of the first and third terms.

The first three options have (d=-8), but (81,72,63,54,\ldots) has (d=-9). Check both value and sign in the options.

The governing concept is the general form of an arithmetic progression. In three consecutive terms written as a, a+d, a+2d, the constant d is the common difference between successive terms. Comparing the given first two terms, 4+d=1, so subtracting 4 from both sides gives d=−3. This result is confirmed by the third term: 4+2d=4+2(−3)=4−6=−2, exactly as required. Therefore option B is correct. Option A has the wrong sign and would produce the sequence 4,7,10. Option C would make the second term 9, and option D would make the second term −1, so neither matches the supplied terms. The negative value simply means that the arithmetic progression decreases by 3 at each step.

The two common differences are (3) and (5), so the sum sequence has (d=3+5=8). In termwise addition, common differences add.

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

Both differences are (5) and (5), so it is an arithmetic progression for every (x). Compare differences before fixing the variable.

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.

(2(x+6)=2x+18) gives (2x+12=2x+18), which is false, so no (x) is possible. Hence (d) is not determined.

An arithmetic progression has the same common difference between every pair of consecutive terms. In option A, the first differences are 6−2=4 and 10−6=4, but the next difference is 15−10=5. Since the difference changes from 4 to 5, the sequence fails the defining condition of an AP. In option B, every displayed difference is 5; in option C, every difference is −4; and in option D, every difference is 0. Thus option A is the only sequence that is not an arithmetic progression. Checking all consecutive differences, rather than only the first one, is essential.

The new terms are (a+1, a+d+2, a+2d+3), and both differences are (d+1). Adding increasing numbers respectively adds (1) to (d).

The first has (d=4) and the second has (d=5), so the difference sequence has (d=4-5=-1). The termwise difference of two arithmetic progressions is also an arithmetic progression.

The first sequence has (d=0) and the second has (d=2), so the new (d=2). Adding a constant sequence does not change (d).

In an arithmetic progression, every consecutive pair has the same common difference. The given consecutive terms are \(-4,s,8,14\). From 8 to 14, the difference is \(14-8=6\), so the common difference must be 6 throughout the progression. Moving one step backward from 8 means subtracting 6, giving \(s=8-6=2\). Therefore, option B is correct.

This can also be checked from the first terms: \(s-(-4)=s+4\) must equal 6, so \(s=2\). Then the sequence becomes \(-4,2,8,14\), whose consecutive differences are all 6. The value 0 would make the first difference 4, and 4 would make it 8; neither agrees with the difference between 8 and 14. Thus 2 is the only suitable value.

The original (d=3); doubling makes (d=6), and subtracting (1) does not change (d). Multiplication changes (d), equal subtraction does not.

The fourth term is 3+3d, so the given condition gives 3+3d=24. Hence 3d=21 and d=7. The second term is 3+d=3+7=10. Therefore option C is correct. The other choices do not satisfy the same value of d: using 8, 9 or 12 as the second term would make the fourth term different from 24.

(12-\frac{3}{2}=\frac{21}{2}), and (-\frac{3}{2}) continues to be subtracted. Match both (a) and (d) together.

For three numbers to be consecutive terms of an arithmetic progression, twice the middle term must equal the sum of the first and third terms. Therefore, 2(3u+2)=(2u−1)+(5u+1). Expanding gives 6u+4=7u, so u=4. Substitution confirms the result: the three terms become 7, 14, and 21, whose consecutive differences are both 7. Hence option D, not option B, is correct. The original marked answer was inconsistent with its own calculation; replacing the selected option ensures that there is one unambiguous correct answer.

When all terms are multiplied by (-\frac{1}{2}), (d) is also multiplied by it, so the new (d=-3). The multiplier applies directly to the common difference too.

The new terms are (81,225,441), and the differences are (144,216), which are not equal. Squaring generally does not preserve an arithmetic progression.