Introduction to APs and common difference.

The first sequence has (d=4) and the second has (d=6), so the difference sequence has (d=4-6=-2). In termwise difference, take the difference of common differences.

From (6+d=11), (d=5), and (6+3d=21) confirms it. When two conditions are given, verify the answer with both.

From (2(x+2)=x+(2x+1)), (2x+4=3x+1), so (x=3) and (d=5-3=2). Decide (d) only after finding (x).

(25-4=21) is split into three equal gaps, so (d=7), and (x+1=11) gives (x=10). In four consecutive terms, finding (d) from first and last terms is fast.

In three consecutive terms, twice the middle term equals the sum of the other two terms. So (2(x+8)=(2x+1)+(3x-4)) gives (x=3).

(2(3a+6)=(5a-2)+(a+18)) gives (a=2). Then the terms (8,12,20) do not have equal differences, so no arithmetic progression is formed.

The consecutive differences are (p,2p,3p), which are equal only when (p=0). In such questions, first write all differences.

Consecutive terms of an arithmetic progression have the same common difference. Here, the difference between 25 and 19 is 6, so the term immediately before 19 is 19 - 6 = 13. Therefore, u = 13. Option 12 is not correct because the difference from 12 to 19 would be 7, not the common difference 6. Exam tip: Find the difference between known consecutive terms and check it in both forward and backward directions.

In an arithmetic progression, the difference between every pair of consecutive terms is constant. The four terms 2, x, y, 20 contain three equal gaps. The total change from the first to the last term is 20 − 2 = 18, so each common difference is d = 18/3 = 6. Therefore x = 2 + 6 = 8 and y = 8 + 6 = 14. Their sum is x + y = 8 + 14 = 22. Equivalently, the terms are 2, 8, 14, 20, which clearly have equal successive differences. Thus option C is correct; the other choices result from using an incorrect gap or adding the wrong terms.

(\frac{5}{8}-\frac{3}{8}=\frac{2}{8}=\frac{1}{4}). When denominators are equal, subtract the numerators.

The original sequence is an arithmetic progression because consecutive terms differ by 11 − 6 = 5 and 16 − 11 = 5. If each original term t is transformed into 2t + 3, the difference between two transformed consecutive terms is (2(t + 5) + 3) − (2t + 3) = 10. The multiplication by 2 scales the common difference from 5 to 10. Adding the same constant 3 to every term shifts the entire sequence but does not change the difference between terms, because the two +3 terms cancel. Therefore the new common difference is 10, so option C is correct.

Adding the same number to all terms does not change the common difference. The original (d=20-24=-4), so the new (d) remains (-4).

The original (d=5), but in reverse order (18,13,8,3,\ldots), (d=-5). Reversing the order changes the sign.

Both consecutive differences are (6) and (6), so it is an arithmetic progression for every (x). Simplify differences before deciding the variable.

The common difference of the original sequence is \(5-9=-4\). Taking the opposite of every term gives \((-9,-5,-1,3,\ldots)\). The difference between consecutive terms is \(-5-(-9)=4\), so the new common difference is \(4\). The value \(-4\) is the common difference of the original sequence, not the new one. Exam tip: when every term is multiplied by \(-1\), the sign of the common difference also changes.

In the first option, every consecutive difference is (\frac{3}{5}). In fraction options, check all differences separately.

(20-2=18) is split into three gaps, so the second term should be (8) and the third (14). This gives (x=5) and (x=\frac{13}{3}), so no single value is possible.

The common difference is found by subtracting a term from the next term. Here, \(1.05-0.60=0.45\) and \(1.50-1.05=0.45\), so the common difference is \(0.45\). Choosing \(0.40\) would not give the correct difference between consecutive terms. Exam tip: Align decimal points before subtracting decimal numbers.

In an arithmetic progression, the common difference is the difference between consecutive terms. Here, \((r-3)-(r-9)=r-3-r+9=6\). Checking the next pair, \((r+3)-(r-3)=6\) as well. Therefore, \(d=6\). Option \(r\) is incorrect because the common difference must not depend on the variable. Exam tip: Find \(d\) by subtracting the first term from the second term.

In an arithmetic progression, twice the middle term equals the sum of the first and third terms. Thus, \(2(4m+2)=3m+(6m-2)\). This gives \(8m+4=9m-2\), so \(m=6\). If 5 is substituted, the two consecutive differences are not equal. Exam tip: for three AP terms, use \(2b=a+c\) directly.