Introduction to APs and common difference.

The common difference between consecutive terms is constant. Here, the difference between 10 and 18 is \(18-10=8\). Therefore, the term immediately before 10 is \(10-8=2\), so \(s=2\). Check: in \(-6,2,10,18\), 8 is added each time. Exam tip: first find the common difference using two known adjacent terms near the missing term.

The original (d=4); doubling makes (d=8), and subtracting the same (1) does not change (d). Multiplication changes (d), equal subtraction does not.

The terms shown have the standard AP form: first term 4, second term \\(4+d\\), third term \\(4+2d\\), and fourth term \\(4+3d\\). If the fourth term is known, first use it to determine the common difference. Then substitute that difference into the expression for the second term.

The fourth term is 31, so \\(4+3d=31\\). Subtracting 4 gives \\(3d=27\\), and dividing by 3 gives \\(d=9\\). The second term is therefore \\(4+d=4+9=13\\). Hence choice B is correct. The value 31 is the fourth term, not the difference, and 15 would result from an incorrect placement of the multiplier on \\(d\\).

(18-\frac{3}{2}=\frac{33}{2}), and (-\frac{3}{2}) continues to be added. Check both (a) and (d) together.

From (2(3u+2)=(2u-3)+(5u-1)), (6u+4=7u-4), so (u=8). Use the twice-middle-term rule to find the variable.

When all terms are multiplied by (-\frac{1}{2}), (d) is also multiplied by it, so the new (d=-5). The multiplier applies directly to the common difference.

The new terms are (16,81,196), and the differences are (65,115), which are not equal. Squaring generally does not preserve an arithmetic progression.

In three consecutive terms of an arithmetic progression, the middle term equals the average of the first and last terms. Therefore h = (6 + 24)/2 = 30/2 = 15. The question asks for the distance from the first term to the middle term, so h − 6 = 15 − 6 = 9. This also follows by observing that the total change 24 − 6 = 18 is divided into two equal common differences, each equal to 9. Hence the sequence is 6, 15, 24, and option B is correct. Option C is the total change, not one common difference.

The original (d=5), and (4n) has (d=4), so the new (d=1). In term-number changes, subtract its difference.

For three quantities to be consecutive terms of an arithmetic progression, twice the middle term must equal the sum of the first and third terms. Here this gives 2(k + 5) = (k − 2) + (2k + 1). Expanding, 2k + 10 = 3k − 1, so k = 11. A direct check gives the terms 9, 16, 23, whose successive differences are 7 and 7, confirming the result. Therefore option D is correct. The earlier option value 9 was not valid because substituting k = 9 gives 7, 14, 19, with unequal differences 7 and 5; it has been corrected to 11.

The new terms are (2,8,16,\ldots), and the differences are (6,8,\ldots), which are not equal. Adding squares of term numbers does not preserve equal difference.

Given \(a+3d-a=36\). Cancelling \(a\) gives \(3d=36\), so \(d=36/3=12\). There are three equal gaps, not four, between the first term \(a\) and the fourth term \(a+3d\). Exam tip: In the \(n\)th term \(a+(n-1)d\) of an AP, count the number of gaps between terms carefully.

(25-4=21) is split into three equal gaps, so the second term should be (11) and the third (18). This makes (x=11) and (2x+1=18) impossible together.

The selected sequence is (5,19,33,\ldots), and its difference is (14). Selecting every second term makes the new (d) twice the original (d).

The difference between consecutive terms is (4). In exams, for linear (a_n), the coefficient of (n) is the common difference.

The differences are not equal, so it is not an AP. In exams, decide by checking consecutive differences, not by looking at the terms only.

Equating differences gives (p+4=3p-8), so (p=6) and (d=10). For three consecutive terms, set second minus first equal to third minus second.

For three consecutive terms of an AP, the middle term is the average of the first and third terms. Hence, \(x=\frac{7+31}{2}=19\). The common difference is \(d=19-7=12\), and this is verified by \(31-19=12\). In option A, the difference from \(7\) to \(18\) is \(11\), but from \(18\) to \(31\) it is \(13\), so it is not an AP. Exam tip: for three consecutive AP terms, use \(2\times\text{middle term}=\text{first term}+\text{third term}\).

Subtracting each term from the next gives (r). In exams, apply the same difference rule even with symbols.

The consecutive difference is the constant (6), so the sequence is an AP. In exams, a constant (a_{n+1}-a_n) identifies an AP.