Introduction to APs and common difference.

Both differences are (3x-4), so the terms form an AP for every real (x). In exams, if both differences are identical expressions, no separate solving is needed.

(\frac{1}{6}-\frac{2}{3}=-\frac{1}{2}), and the same difference continues. In exams, use common denominators with negative fractions.

The first terms are (2,\frac{3}{2},\frac{4}{3},\ldots), and the differences change. In exams, test fractional forms in (n) by differences.

In the reversed order, each step subtracts (d), so the new difference is (-d). In exams, reversing the order can change the sign of the difference.

The coefficient of (n) is (-5), so it is the common difference. In exams, first check the coefficient of (n) in a linear term.

The consecutive differences are not equal, so it is not an AP. In exams, test constancy of differences, not just the visible pattern.

Equal differences give (-3t+9=-2t+11), so (t=-2) and (d=15). In exams, subtract first from second and second from third.

The middle term is the average of the extremes, so (z=\frac{-17+23}{2}=3) and (d=20). In exams, handle signs carefully when averaging negatives.

Find the difference between consecutive terms rather than looking at the constant part of each expression. The first difference is (2p + 2q) − (2p − q) = 3q. The next difference is (2p + 5q) − (2p + 2q) = 3q, and the following one is (2p + 8q) − (2p + 5q) = 3q. Since the difference is constant, the sequence is an AP with d = 3q. Thus option D is correct. The 2p terms cancel during subtraction, so 2p is not the common difference. q and 2q also do not equal the actual change between consecutive terms.

The difference depends on (n), so it is not constant for all (n). In exams, if (n) remains in the difference, it is not an AP.

The coefficient of (n) is (\frac{5}{4}). In exams, the same rule works for linear formulas with fractions.

(a_{10}-a_4=6d=-18), so (d=-3). In exams, divide the total change by the gap in term numbers.

Equating differences gives (x+5=3x-8), so (x=\frac{13}{2}) and (d=\frac{23}{2}). In exams, do not reject a fractional answer too quickly.

In that option, each step increases by (0.75). In exams, use consecutive difference as the test even with decimals.

Each step decreases by (3.5^\circ), so (d=-3.5^\circ). In exams, use a negative sign for a decreasing sequence.

Putting (n=1) gives (a_1=c), and each next term subtracts (r). In exams, compare with the form (a+d(n-1)).

The terms become (\sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3}). In exams, simplify radicals before finding differences.

Writing the terms as (p,p+d,p+2d,p+3d) gives (p+s=q+r). In exams, verify four-term relations symbolically.

In an AP, consecutive differences are equal, so (b-a=c-b). In exams, the same condition can be written as (2b=a+c).

In an AP, the common difference d is added to each term to get the next term. Here d = -2, so starting from 11 gives 9, 7, and 5. Option B has a common difference of +2, so it is incorrect. Exam tip: a negative d means that the terms decrease successively.