Introduction to APs and common difference.

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.

Multiplying by (\frac{1}{3}) changes the difference from (-9) to (-3), and adding (5) does not change it. In exams, separate the effects of addition and multiplication.

The new sequence jumps three original terms each time, so its difference is (3d=12). In exams, count the gap between selected term numbers.

The new common difference is 6, so D is correct. In the original finite AP, moving forward one place changes a term by d=−6. When the terms are written backward, moving forward in the reversed list means moving backward in the original list; therefore each change is the opposite, −d=−(−6)=6. A, −12, doubles the magnitude incorrectly. B, −6, is the old difference and does not account for reversal. C, 0, would mean all terms are equal, which is not true when the original difference is −6. D, 6, is the correct sign reversal. For example, 20,14,8 becomes 8,14,20, whose differences are 6. Memory cue: reversing a finite AP changes d to −d, not its magnitude to a new value.

The condition (2r^2=r+r^3) gives (r(r-1)^2=0), so the nonzero value is (1). In exams, always apply the nonzero condition.

For three AP terms, (2\beta=\alpha+\gamma), so (\beta=7). In exams, treat the middle term as the average of the extremes.

(a_n=(m+1)n), so the coefficient of (n) is (m+1). In exams, combine like terms before identifying the linear coefficient.

The consecutive differences keep increasing, so they are not constant. In exams, a famous pattern need not be an AP.

(\frac{5}{8}-\frac{7}{8}=-\frac{2}{8}=-\frac{1}{4}). In exams, with equal denominators, subtract numerators directly.

The middle-term condition gives (x^2-4=x^2), which is impossible. In exams, also check that denominators are nonzero.

The consecutive differences are not equal, so it is not an AP. In exams, do not confuse a multiplicative pattern with an AP.

The common difference is second term minus first term, so (3\lambda+2-(\lambda-4)=2\lambda+6). In exams, handle the minus sign before brackets carefully.

The initial terms are (1,5,5,9,\ldots), so the difference is not constant. In exams, test terms containing ((-1)^n) by substituting a few values.

The distance increases by (12) kilometres each time. In word problems, still find the same consecutive difference.

The differences are (k+1,3k-3,3k-3), and equality gives (k=2). In exams, check all consecutive differences for four terms.

Here (c_n=4) for every (n), so its common difference is (0). In exams, treat the sequence of differences as a separate sequence.

Checking only the first three terms is not enough for the whole sequence. In exams, the full condition must involve every (n) or all consecutive terms.

Find the difference between consecutive terms: \(10-3=7\), \(17-10=7\), \(24-17=7\), and \(31-24=7\). Therefore, this is an AP with common difference \(7\) days. \(14\) days represents two weeks, but each successive date here is only 7 days later. Exam tip: To find the common difference, subtract the first term from the second term.

The terms are (\frac{1}{3},\frac{1}{5},\frac{1}{7},\ldots), and the differences change. In exams, test fractions with changing denominators by differences.

The new sequence jumps two original terms each time, so the difference is (2d). In exams, multiply (d) by the term-number jump.

Moving two terms ahead changes the value by (2d), so (2d=18) and (d=9). In exams, count the number of jumps.

There are (5) intervals between (a_7) and (a_2), so (5d=35) and (d=7). In exams, subtract term numbers to get intervals.

The sum sequence has difference (3+(-5)=-2). In exams, for sums of corresponding terms, add the differences too.