Introduction to APs and common difference.

Multiplying by (\frac{1}{3}) changes the difference from (-9) to (-3), and adding (5) does not change it. In exams, separate the effects of addition and multiplication.

The new sequence jumps three original terms each time, so its difference is (3d=12). In exams, count the gap between selected term numbers.

Reversing changes each step to the opposite sign. In exams, reverse order changes (d) to (-d).

The condition (2r^2=r+r^3) gives (r(r-1)^2=0), so the nonzero value is (1). In exams, always apply the nonzero condition.

For three AP terms, (2\beta=\alpha+\gamma), so (\beta=7). In exams, treat the middle term as the average of the extremes.

(a_n=(m+1)n), so the coefficient of (n) is (m+1). In exams, combine like terms before identifying the linear coefficient.

The consecutive differences keep increasing, so they are not constant. In exams, a famous pattern need not be an AP.

(\frac{5}{8}-\frac{7}{8}=-\frac{2}{8}=-\frac{1}{4}). In exams, with equal denominators, subtract numerators directly.

The middle-term condition gives (x^2-4=x^2), which is impossible. In exams, also check that denominators are nonzero.

To test three symbolic terms for an arithmetic progression, calculate the two consecutive differences. The first is (a + b) − (a − 2b) = 3b. The second is (a + 4b) − (a + b) = 3b. These differences are identical for all values of a and b, so the three terms are always consecutive terms of an AP with common difference d = 3b. Therefore option A is correct. The quantity a cancels in both subtractions, so it cannot be the common difference. No extra condition such as a = b is required, and the equal differences disprove the claim that the terms are not in AP.

The consecutive differences are not equal, so it is not an AP. In exams, do not confuse a multiplicative pattern with an AP.

The common difference is second term minus first term, so (3\lambda+2-(\lambda-4)=2\lambda+6). In exams, handle the minus sign before brackets carefully.

The initial terms are (1,5,5,9,\ldots), so the difference is not constant. In exams, test terms containing ((-1)^n) by substituting a few values.

The common difference is the amount added to obtain each next term. Subtracting the first term from the second gives (x + √5) − x = √5. The next subtraction gives (x + 2√5) − (x + √5) = √5, and the third gives (x + 3√5) − (x + 2√5) = √5. Thus the sequence is an arithmetic progression with common difference √5, so option B is correct. The variable x is the initial or fixed part and cancels when consecutive terms are subtracted. Option C doubles the actual step, while option D is a term, not the difference.

The distance increases by (12) kilometres each time. In word problems, still find the same consecutive difference.

The differences are (k+1,3k-3,3k-3), and equality gives (k=2). In exams, check all consecutive differences for four terms.

Here (c_n=4) for every (n), so its common difference is (0). In exams, treat the sequence of differences as a separate sequence.

The governing property is that consecutive terms of an AP have equal differences. Therefore, m − 14 must equal 50 − m. Solving gives 2m = 64, so m = 32. The common difference is then d = m − 14 = 32 − 14 = 18; it also agrees with 50 − 32 = 18. Hence option B is correct. Option A incorrectly uses 30 as the middle term and consequently gets a difference of 16. Options C and D place the middle term too high, so their two successive differences are not equal: 20 and 16 for C, and 22 and 14 for D. The arithmetic-mean rule, m = (14 + 50)/2, provides the same result.

Checking only the first three terms is not enough for the whole sequence. In exams, the full condition must involve every (n) or all consecutive terms.

The governing concept is the common difference of an arithmetic progression, defined as the difference between any term and the immediately preceding term. Subtract the first term from the second: d = 72 − 81 = −9. Checking the next pairs gives 63 − 72 = −9 and 54 − 63 = −9, confirming that the sequence is an AP with d = −9. Therefore option B is correct. The negative sign is essential because the terms decrease as the sequence advances. Option A gives the magnitude but loses the direction, option C doubles the actual change, and option D is merely a term of the sequence rather than a difference. Thus the constant signed difference is −9.