Introduction to APs and common difference.

Subtracting each term from the next gives (r). In exams, apply the same difference rule even with symbols.

The consecutive difference is the constant (6), so the sequence is an AP. In exams, a constant (a_{n+1}-a_n) identifies an AP.

Here the difference depends on (n), so it is not constant. In exams, a quadratic (a_n) usually does not form an AP.

In an AP, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_9-a_5=(9-5)d\), so \(34-18=4d\). Hence \(16=4d\) and \(d=4\). Option 16 is the difference between the two given terms, not the common difference. Exam tip: find the difference in term numbers first, then divide the term difference by it.

Both differences are (2k-3), so it forms an AP for every real (k). In exams, simplify both differences first.

An arithmetic progression must have the same difference between every two consecutive terms. The required difference here is \(\frac14\), so subtract each term from the next and check whether the result is always \(\frac14\). Option D follows the pattern of adding one-fourth at every step, so it is the correct sequence.

For option D, \(\frac34-\frac12=\frac14\), \(1-\frac34=\frac14\), and \(\frac54-1=\frac14\). All consecutive differences are equal. In option A, the differences are not all equal; option B also has changing differences, and option C doubles irregularly. Therefore only option D is an AP with common difference \(\frac14\).

Each next angle decreases by (30^\circ), so the difference is (-30^\circ). In exams, keep the unit in the answer.

The coefficient of (n) is (-\frac{2}{5}), which is the common difference. In exams, identify the form (a_n=\alpha n+\beta).

Putting (n=1) gives the first term (11), and each step adds (-3). In exams, recognize the form (a+d(n-1)).

The terms become (\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2}), so the difference is (\sqrt{2}). In exams, simplify radicals first.

Equal differences give (5-m=3m-7), so (m=3) and (d=2). In exams, be careful with signs in terms containing variables.

In an AP, the middle term is the average of the two extremes, so (2b=a+c). In exams, this is the fastest check for three terms.

In an arithmetic progression, the difference between every pair of consecutive terms must be the same. In option D, the consecutive differences are 3, 3, and 4, which are not equal; hence it is not an AP. In options A, B, and C, the common differences are 3, 4, and 0 respectively, all constant. Exam tip: check every consecutive difference, not only the first two.

The amount increases by (15) rupees each time, so (d=15). In word problems, still look for the consecutive difference.

The condition (2k^2=k+k^3) gives (k(k-1)^2=0), and the nonzero value is (1). In exams, do not ignore conditions like nonzero.

The middle-term condition leads to an impossible equation, so there is no valid (p). In exams, also check that denominators are nonzero.

Both differences are (h), so it is an AP for every real (h). In exams, remember that (h=0) gives a valid constant AP.

(3n+1) is linear and has difference (3). In exams, a linear form with constant coefficient gives an AP.

In an AP, \(a_3=a_1+2d\). Therefore, \(a_3-a_1=2d=14\), so \(d=7\). The value \(-7\) would apply only if the given difference were \(-14\). Exam tip: use \(a_m-a_n=(m-n)d\) and carefully count the difference between the term numbers.

In an AP, (a_2+a_4=2a_3) is automatically true, so (d) is not found. In exams, check whether the information is new or just an identity.

The definition of an AP is based on a constant consecutive difference. In definition-based questions, use this test.

Every consecutive difference is (0), so it is a constant AP. In exams, treat (d=0) as a valid common difference.

The middle-term condition gives (x(x+2)=(x+1)^2), which is impossible. In exams, check validity after forming the equation.

(a_{n+1}-a_n=\alpha), so the common difference is (\alpha). In exams, this is the key shortcut for a linear sequence.

Taking the terms as (a,a+d,a+2d,a+3d), both sides become (2a+3d). In exams, test relations by writing symbolic AP terms.