Introduction to APs and common difference.

The consecutive differences are (r,3r,5r), which are equal only when (r=0). In such questions it is necessary to write all consecutive differences.

The original (d=-4), and in reverse order (6,10,14,18,\ldots), (d=4). Reversing the order changes the sign of (d).

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

The first sequence has (d=-6) and the second has (d=5), so the difference sequence has (d=-6-5=-11). In termwise subtraction, the common differences also subtract.

\(-1-\left(-\frac{7}{4}\right)=\frac{3}{4}\), and the next difference is also \(\frac{3}{4}\). Be careful while subtracting negative fractions.

In an arithmetic progression, the common difference d is the difference between consecutive terms. Here, d = 1.35 - 0.80 = 0.55. Checking further, 1.90 - 1.35 = 0.55 and 2.45 - 1.90 = 0.55. Therefore, the correct answer is 0.55; 0.50 is not the actual difference between the decimal terms. Exam tip: Align decimal points while subtracting decimals.

In an arithmetic progression, the common difference is the difference between consecutive terms. Here, \((u-2)-(u-8)=6\) and \((u+4)-(u-2)=6\). Thus, 6 is added to each successive term, so \(d=6\). \(u+6\) may be a next term, not the common difference. Exam tip: quickly check \(d=a_2-a_1\).

From the third to the seventh term there are (4) gaps, so (4d=28) and (d=7). Convert term distance into number of gaps.

From the second to the fifth term there are (3) gaps, so (3d=-18) and (d=-6). In a decreasing progression, (d) remains negative.

The new terms are (a+2, a+d+4, a+2d+6), and both differences are (d+2). The difference (2) of the added numbers is also added to (d).

The original (d=6), and (n) has difference (1), so the new (d=6-1=5). When subtracting term number, its difference is also subtracted.

An arithmetic progression must have the same difference between every consecutive pair of terms. Matching only the first two differences does not prove that a sequence is an AP; later terms must also be checked. A single changed difference is enough to disqualify the sequence.

In choice A, the first difference is \\(9-4=5\\), and the second is \\(14-9=5\\). However, the third difference is \\(20-14=6\\), not 5. Therefore the difference is not constant, so choice A is not an arithmetic progression. In B the difference is always 5, in C it is always \\(-6\\), and in D it is always 0. Thus A is the only correct choice.

Both consecutive differences are (6) and (6), so it is an arithmetic progression for every (x). Simplify differences before fixing the variable.

(2(x+9)=2x+18) gives an identity, so it is an arithmetic progression for every (x). Therefore no single value of (x) is obtained.

The original (d=6), and multiplying by (-\frac{1}{3}) gives new (d=-2). The multiplier applies directly to the common difference too.

The new terms are (9,49,121), and the differences are (40,72), which are not equal. Squaring generally does not preserve an arithmetic progression.

The fourth term is 7+3d, and it is given as 34. Therefore 7+3d=34, which gives 3d=27 and d=9. The second term is 7+d, so its value is 7+9=16. Hence option D is correct. The sequence becomes 7, 16, 25, 34, and each consecutive difference is 9, confirming that it is an arithmetic progression. A common mistake is to divide the total change 34−7 by 4 instead of by 3; there are three equal gaps between four terms, so division by 3 is required.

(15-\frac{5}{2}=\frac{25}{2}), and (-\frac{5}{2}) continues to be added. Check both (a) and (d) conditions together.

From (2(4v+1)=(3v-4)+(6v-1)), (8v+2=9v-5), so (v=7). Use the twice-middle-term rule to find the variable.

The original (d=3), and (4n) has (d=4), so the new (d=3-4=-1). In term-number based changes, change the differences too.