Introduction to APs and common difference.

Equating differences gives ((3x+10)-(5x+2)=(x+18)-(3x+10)), which is an identity, so (d=8-2x). Therefore it is an arithmetic progression for every (x), but (d) is not fixed.

Adding the same (5) to all terms does not change (d), so (d=-4) remains. Equal addition or subtraction keeps the common difference unchanged.

The original common difference is (5), so after multiplying by (3), the new (d=15). When all terms are multiplied by the same factor, (d) is multiplied by that factor too.

The original (d=7), and in reverse order (23,16,9,2,\ldots), (d=-7). Reversing the order changes the sign of the common difference.

Both differences are (x-2) and (x-2), so they are equal for every (x). Simplify and compare differences in algebraic terms.

The original (d=-4), and the opposite terms form (-7,-3,1,5,\ldots), whose (d=4). Multiplying all terms by (-1) changes the sign of (d).

In the first option, (1-\frac{1}{6}=\frac{5}{6}) and (\frac{11}{6}-1=\frac{5}{6}). Check every difference separately in options.

The total difference (14-2=12) is split into three equal gaps, so (d=4), and the second term should be (6), hence (x=5). Splitting the total difference into equal parts is useful.

The common difference is found by subtracting a term from the next term: \(1.20-0.75=0.45\). Checking further, \(1.65-1.20=0.45\) and \(2.10-1.65=0.45\). Hence, this is an AP with common difference 0.45. Choosing 0.40 does not give the actual difference between consecutive terms. Exam tip: Align decimal points before subtracting decimal numbers.

In an arithmetic progression, the common difference is the difference between consecutive terms. Here, \((r-1)-(r-6)=5\) and \((r+4)-(r-1)=5\). Since both differences are equal, \(d=5\). \(r+5\) may describe a term expression, but it is not the common difference. Exam tip: subtract any term from the next term to find \(d\).

From (2(3a+1)=2a+(5a-1)), (6a+2=7a-1), so (a=3). The twice-middle-term rule works quickly for three terms.

The original (d=4), and the difference of (2n) is (2), so the new (d=4-2=2). In term-number transformations, subtract the difference of the change.

From the second to the fifth term there are (3) gaps, so (3d=18) and (d=6). Convert distance between terms into number of gaps.

There are (3) gaps between the fourth and first terms, so (3d=-15) and (d=-5). Keep (d) negative for a decreasing progression.

The original (d=3), and term numbers (1,2,3,\ldots) have difference (1), so the new (d=4). In this transformation, add the two common differences.

The consecutive differences are (2q,2q,2q), so it is an arithmetic progression for every (q). Equal algebraic differences are also valid.

The original (d=-3), and multiplying by (-2) gives new (d=6). If the multiplier is negative, the sign of (d) can also change.

From (2(2x+3)=x+(4x+1)), (4x+6=5x+1), so (x=5) and (d=13-5=8). Finding the variable first is necessary.

\(-\frac{5}{6}-\left(-\frac{3}{2}\right)=\frac{2}{3}\), and the next difference is also \(\frac{2}{3}\). Be careful while subtracting negative fractions.

From (2(2m+5)=(m+2)+(4m+1)), (4m+10=5m+3), so (m=7). Use the twice-middle-term rule for three terms.