Introduction to APs and common difference.

Correct answer: A. (4)

From (2(2m+5)=(m+2)+(4m+1)), (4m+10=5m+3), so (m=7). Use the twice-middle-term rule for three terms.

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

The original (d=-5), so after dividing by (4), the new (d=-\frac{5}{4}). Dividing all terms by the same number divides (d) by that number.

Correct answer: B. 9

In an arithmetic progression, the difference between consecutive terms is constant. Therefore, for three terms, the middle term b is the average of a and c: \(b=\frac{a+c}{2}=\frac{4+22}{2}=13\). Hence, \(b-a=13-4=9\). Option 11 can result from using an incorrect difference instead of halving \(c-a=22-4\). Exam tip: For three consecutive AP terms, the middle term is always the average of the first and third terms.

Correct answer: D. (81,72,63,54,\ldots)

The first three options have (d=-8), but (81,72,63,54,\ldots) has (d=-9). Check both value and sign in the options.

Correct answer: C. (8)

The two common differences are (3) and (5), so the sum sequence has (d=3+5=8). In termwise addition, common differences add.

Correct answer: A. (-7)

The first sequence has (d=-3) and the second has (d=4), so the difference sequence has (d=-3-4=-7). In termwise subtraction, subtract the common differences too.

Correct answer: C. True for every (x)

Both differences are (5) and (5), so it is an arithmetic progression for every (x). Compare differences before fixing the variable.

Correct answer: C. (-4)

The original (d=-3), and term numbers have (d=1), so the new (d=-3-1=-4). When subtracting term number, its difference is also subtracted.

Correct answer: D. (6)

(2(x+6)=2x+18) gives (2x+12=2x+18), which is false, so no (x) is possible. Hence (d) is not determined.

Correct answer: C. (d+1)

The new terms are (a+1, a+d+2, a+2d+3), and both differences are (d+1). Adding increasing numbers respectively adds (1) to (d).

Correct answer: A. Arithmetic progression with (d=-1)

The first has (d=4) and the second has (d=5), so the difference sequence has (d=4-5=-1). The termwise difference of two arithmetic progressions is also an arithmetic progression.

Correct answer: B. (2)

The first sequence has (d=0) and the second has (d=2), so the new (d=2). Adding a constant sequence does not change (d).

Correct answer: B. (2)

In an arithmetic progression, every consecutive pair has the same common difference. The given consecutive terms are \(-4,s,8,14\). From 8 to 14, the difference is \(14-8=6\), so the common difference must be 6 throughout the progression. Moving one step backward from 8 means subtracting 6, giving \(s=8-6=2\). Therefore, option B is correct.

This can also be checked from the first terms: \(s-(-4)=s+4\) must equal 6, so \(s=2\). Then the sequence becomes \(-4,2,8,14\), whose consecutive differences are all 6. The value 0 would make the first difference 4, and 4 would make it 8; neither agrees with the difference between 8 and 14. Thus 2 is the only suitable value.

Correct answer: C. (6)

The original (d=3); doubling makes (d=6), and subtracting (1) does not change (d). Multiplication changes (d), equal subtraction does not.

Correct answer: A. (12, \frac{21}{2}, 9, \frac{15}{2},\ldots)

(12-\frac{3}{2}=\frac{21}{2}), and (-\frac{3}{2}) continues to be subtracted. Match both (a) and (d) together.

Correct answer: B. (-3)

When all terms are multiplied by (-\frac{1}{2}), (d) is also multiplied by it, so the new (d=-3). The multiplier applies directly to the common difference too.

Correct answer: C. No because consecutive differences are not equal

The new terms are (81,225,441), and the differences are (144,216), which are not equal. Squaring generally does not preserve an arithmetic progression.

Correct answer: B. \(8\)

In an arithmetic progression with three terms, the middle term equals the average of the first and third terms. Thus, \(h=\frac{4+20}{2}=12\). Therefore, \(h-4=12-4=8\). Note that \(12\) is the value of \(h\), whereas the question asks for \(h-4\). Exam tip: for three AP terms, use middle term = average of the end terms.

Correct answer: A. (1)

The original (d=4), and (3n) has (d=3), so the new (d=1). In term-number changes, subtract its common difference.

Correct answer: B. (6)

In an arithmetic progression, the middle term is the average of the extremes, so (2(5k+1)=(3k-2)+(8k-3)). For three terms, this is the fastest exam rule.

Correct answer: B. It is never an arithmetic progression

The consecutive differences are (2d) and (3d), which cannot be equal when (d \neq 0). In exams, compare consecutive differences directly.

Correct answer: B. 46

In an arithmetic progression, consecutive terms have the same difference. There are 4 gaps between the first and fifth terms, so the common difference is \(d=\frac{39-7}{4}=8\). Thus, \(x=7+8=15\) and \(y=23+8=31\). Therefore, \(x+y=15+31=46\). Option 42 can result from counting the gaps incorrectly. Exam tip: between \(n\) terms, there are always \(n-1\) gaps.

Correct answer: B. 13

In an arithmetic progression, the differences between consecutive terms are equal. Therefore,
\((2p+1)-(p-4)=(4p-2)-(2p+1)\).
This gives \(p+5=2p-3\), so \(p=8\). The common difference is then \((2p+1)-(p-4)=17-4=13\). Values such as 15 do not equal the consecutive difference of the given terms. Exam tip: for three terms in an AP, set twice the middle term equal to the sum of the first and third terms.

Correct answer: C. 2

In an arithmetic progression, the differences between consecutive terms must be equal. Here, the first difference is \(q^2+q-q^2=q\), and the second difference is \(q^2+3q-2-(q^2+q)=2q-2\). Thus, \(q=2q-2\), which gives \(q=2\). Therefore, option 2 is correct. For instance, when \(q=1\), the differences are 1 and 0, so the terms are not in AP. Exam tip: for three terms \(a,b,c\) in AP, you may also use \(2b=a+c\).

Correct answer: A. The differences between consecutive terms are 5, 5 and 6, which are not equal.

In an AP, the difference between every pair of consecutive terms must be constant. Here, 9−4=5 and 14−9=5, but 20−14=6. So it is not an AP. Exam tip: check all consecutive differences.