Introduction to APs and common difference.

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.

The middle term is (y=\frac{-11+5}{2}=-3), and (d=8). In exams, be careful while averaging negative numbers.

Here the common difference is (-3), so the terms are decreasing. In exams, a decreasing AP has negative (d).

Equating differences gives (-2q-6=-2q+14), which is impossible. In exams, cancellation of the variable can produce a contradiction.

The signs alternate and the difference does not remain the same. In exams, test alternating-sign sequences by differences only.

Use the standard nth-term form of an arithmetic progression, aₙ = a₁ + (n − 1)d. The given expression is aₙ = 12 − 7(n − 1), which can be read directly as a₁ + (n − 1)d with a₁ = 12 and d = −7. A quick check confirms this: when n = 1, a₁ = 12 − 7(0) = 12; when n = 2, a₂ = 5, so a₂ − a₁ = 5 − 12 = −7. Thus option C is correct. Option A loses the negative sign, option B incorrectly treats the second term as the first, and option D reverses the two quantities.

The first term is the first number in each sequence, and the common difference is the change from one term to the next. In option A, the first term is 6, while 12 − 6 = 6, 18 − 12 = 6, and 24 − 18 = 6. Thus d = 6, equal to the first term. In option B, the difference is 4, not 6. In option C, the difference is 0, and in option D, the difference is −6. Therefore only option A satisfies a₁ = d = 6. The sequence in A is an increasing AP whose terms rise by exactly its initial value.

The common difference is obtained by subtracting a term from the following term. Using the first two terms, d = 0.02 − 0.20 = −0.18. The next subtraction confirms the pattern: −0.16 − 0.02 = −0.18, and −0.34 − (−0.16) = −0.18. Therefore the sequence is an AP with common difference −0.18, so option A is correct. The negative sign means that the terms decrease by 0.18 each time. Option B has the wrong sign, option C results from an incorrect decimal subtraction, and option D is merely one of the terms rather than the change between terms.

Multiplying terms by (2) multiplies the difference by (2), so it becomes (8). In exams, adding or subtracting a constant does not change the difference, but multiplication does.

Adding the same number to every term does not change consecutive differences. In exams, a uniform translation leaves the difference unchanged.

Multiplication scales all differences by the same factor, so the new difference is (-3d). In exams, apply the transformation to the difference.

Subtracting the same number from every term does not change the difference. In exams, treat uniform subtraction as changing only the first term.

For three consecutive terms of an arithmetic progression, the middle term is the arithmetic mean of the other two, so a + c = 2b. Equivalently, the three terms can be written as b − d, b, and b + d. Their sum is (b − d) + b + (b + d) = 3b because the −d and +d cancel. Given a + b + c = 27, we obtain 3b = 27, hence b = 9. Therefore option C is correct. The values 7, 8, and 10 do not satisfy the required sum relation b = 27/3; the answer does not depend on knowing a or c separately.

The consecutive differences are (v) and (v), so they are in AP. In exams, recognize the balanced form (u-v,u,u+v) quickly.