Introduction to APs and common difference.

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.

Each next term adds (4), so the common difference is (4). In exams, a variable constant part does not change the difference.

The terms are (0,0,2,6,\ldots), and the differences (0,2,4,\ldots) change. In exams, expand the product or test initial terms.

Both consecutive differences are (3a+2b), so the terms are in AP. In exams, also check equality of the two differences.

Multiplication makes every difference (t) times as large. In exams, apply scaling directly to the common difference.

In an AP, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_6-a_3=(6-3)d\). Hence \(25-4=3d\), so \(21=3d\) and \(d=7\). Option 21 is the difference between the given terms, not the common difference. Exam tip: when subtracting two given terms, also use the difference between their subscripts.

Each term is (2.5) less than the previous one, so (d=-2.5). In exams, do not forget the negative sign in decreasing decimal sequences.

(a_5-a_2=3d=15), so (d=5). In exams, even for a linear form, (d) can be found from term differences.

There are (4) equal gaps between the first and fifth terms, so (d=\frac{19-3}{4}=4). In exams, divide the total difference between distant terms by the number of gaps.

In an arithmetic progression, the common difference is the constant amount added to obtain each successive term. Subtracting consecutive terms gives 8 − 5 = 3, 11 − 8 = 3, and 14 − 11 = 3. Since the same value is obtained each time, the common difference is 3. Therefore, option B is correct; 2, 4, and 5 do not represent the change between consecutive terms.