Introduction to APs and common difference.

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 has one constant difference between every pair of consecutive terms. For option A, 5 − 9 = −4, 1 − 5 = −4, and −3 − 1 = −4, so every step has the required common difference. Option B has differences −5, −4, and −4; option C has differences +4 throughout, so it is an AP but not with difference −4; option D has differences −4 and −5, so its difference is not constant. Therefore option A is the only sequence satisfying both conditions: it is an AP and its common difference is −4.

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.

An arithmetic progression can be written in the form a_n=a_1+(n-1)d, so the coefficient of n in a linear expression for its terms equals the common difference. Here a_n=7-0.5n. Increasing n by 1 changes the term by -0.5: a_{n+1}-a_n=[7-0.5(n+1)]-[7-0.5n]=-0.5. Thus the sequence decreases by one-half at every step, and its common difference is -0.5. Option D is correct. Option B gives the magnitude but misses the negative sign, while options A and C confuse the constant or the coefficient with the required difference.

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.

The governing concept is that an arithmetic progression has a constant common difference, found by subtracting one term from the next. Here, (3a + 2b) − 3a = 2b. Also, (3a + 4b) − (3a + 2b) = 2b, and (3a + 6b) − (3a + 4b) = 2b. Since the same quantity is added at every step, the sequence is an AP with common difference 2b. Therefore, option C is correct. The terms 3a and a are not the difference: 3a is the fixed algebraic part, while a and b alone do not represent the change between consecutive terms.

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.

For three consecutive terms of an arithmetic progression, the two consecutive differences must be equal. The first difference is (2x + 1) − (x − 2) = x + 3. The second difference is (4x − 3) − (2x + 1) = 2x − 4. Equating them gives x + 3 = 2x − 4, so x = 7. Substituting this value, the first difference is 7 + 3 = 10, and the second difference is 14 − 4 = 10 as well. Hence d = 10 and option C is correct. The other choices fail because their values of x do not make the two differences equal.