Introduction to APs and common difference.

Correct answer: B. It is an AP and (d=4)

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

Correct answer: C. No, the consecutive difference is not constant

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.

Correct answer: A. (3a+2b)

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

Correct answer: C. (t(b-a))

The direct answer is option C,
\(t(b-a)\). Since
\((a,b,c)\) is an AP, its common difference is
\(b-a"); equivalently,
\(c-b=b-a\). After multiplying every term by
\(t"), the new terms are
\(ta,tb,tc"). Their consecutive difference is
\(tb-ta=t(b-a)"). The next difference is
\(tc-tb=t(c-b)=t(b-a)"), so the result is consistent. Option A,
\(b-a"), is the old difference and misses the scaling by t. Option B,
\(t"), is only the multiplier, not generally the new difference. Option C is correct. Option D,
\(a+t"), adds t to one term and does not describe multiplying all terms. If t is negative, the formula still works and reverses the direction as appropriate. Exam cue: multiplying all AP terms by t multiplies d by t.

Correct answer: A. 7

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.

Correct answer: B. (-2.5)

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.

Correct answer: C. (5)

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

Correct answer: B. (x=7,y=15,d=4)

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.