Introduction to APs and common difference.

Correct answer: B. (y=-3,d=8)

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

Correct answer: C. (20,17,14,11,\ldots)

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

Correct answer: D. No, no (q)

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

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

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

Correct answer: D. (8)

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.

Correct answer: A. It remains the old common difference

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

Correct answer: B. (-3d)

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

Correct answer: B. (7)

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

Correct answer: A. They are in AP and (d=v)

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

Correct answer: C. Every real (x)

Both differences are (3x-4), so the terms form an AP for every real (x). In exams, if both differences are identical expressions, no separate solving is needed.

Correct answer: B. (-\frac{1}{2})

(\frac{1}{6}-\frac{2}{3}=-\frac{1}{2}), and the same difference continues. In exams, use common denominators with negative fractions.

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

The first terms are (2,\frac{3}{2},\frac{4}{3},\ldots), and the differences change. In exams, test fractional forms in (n) by differences.

Correct answer: B. (-d)

In the reversed order, each step subtracts (d), so the new difference is (-d). In exams, reversing the order can change the sign of the difference.

Correct answer: A. (-5)

The coefficient of (n) is (-5), so it is the common difference. In exams, first check the coefficient of (n) in a linear term.

Correct answer: D. No, the differences are (7,19,37,\ldots)

The consecutive differences are not equal, so it is not an AP. In exams, test constancy of differences, not just the visible pattern.

Correct answer: C. (t=-2,d=15)

Equal differences give (-3t+9=-2t+11), so (t=-2) and (d=15). In exams, subtract first from second and second from third.

Correct answer: D. (z=3,d=20)

The middle term is the average of the extremes, so (z=\frac{-17+23}{2}=3) and (d=20). In exams, handle signs carefully when averaging negatives.

Correct answer: D. Never

The difference depends on (n), so it is not constant for all (n). In exams, if (n) remains in the difference, it is not an AP.

Correct answer: A. (\frac{5}{4})

The coefficient of (n) is (\frac{5}{4}). In exams, the same rule works for linear formulas with fractions.

Correct answer: C. (-3)

The direct answer is option C,
\(-3\). In an AP, the change between two terms equals the number of equal steps multiplied by the common difference. From the fourth term to the tenth term there are
\(10-4=6\) steps. The total change is
\(a_{10}-a_4=-6-12=-18\). Therefore
\(6d=-18\), so dividing by 6 gives
\(d=-3\). Option A,
\(-6\), forgets that there are six steps. Option B,
\(-4\), does not satisfy
\(6d=-18\). Option C is correct. Option D,
\(3\), has the wrong sign: a term falls from 12 to -6, so the common difference must be negative. Memory cue: subtract term values, count the gap in indices, then divide.

Correct answer: A. (x=\frac{13}{2},d=\frac{23}{2})

Equating differences gives (x+5=3x-8), so (x=\frac{13}{2}) and (d=\frac{23}{2}). In exams, do not reject a fractional answer too quickly.

Correct answer: D. (2.5,3.25,4,4.75)

In that option, each step increases by (0.75). In exams, use consecutive difference as the test even with decimals.

Correct answer: C. (-3.5^\circ)

Each step decreases by (3.5^\circ), so (d=-3.5^\circ). In exams, use a negative sign for a decreasing sequence.

Correct answer: D. (a_1=c,d=-r)

The direct answer is D: \(a_1=c\) and \(d=-r\). Compare the formula with the standard arithmetic-progression form \(a_n=a_1+(n-1)d\). Here \(a_n=c-(n-1)r=c+(n-1)(-r)\). Therefore the first term is c and the common difference is -r. Direct checking gives \(a_1=c-(1-1)r=c\). Also, subtracting consecutive terms gives \(a_{n+1}-a_n=-r\), confirming the difference. Option A incorrectly says the first term is c-r and gives a positive difference. Option B has the correct first term but wrong sign for the difference. Option C swaps the roles of c and r. Option D matches both results. Memory cue: the sign before r is the sign of the common difference.

Correct answer: A. It is an AP and (d=\sqrt{3})

The terms become (\sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3}). In exams, simplify radicals before finding differences.