Introduction to APs and common difference.

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

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.

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.

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

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

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.

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

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.

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

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).

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.

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).

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.

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

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

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.

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.

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.

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

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.

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

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.

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.

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 \(2b=a+c\).

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.