Introduction to APs and common difference.

This is an arithmetic progression because \(6\) is added to each successive term. Here, the first term is \(a=-18\) and the common difference is \(d=6\). Therefore, \(a_6=a+5d=-18+5\times6=12\). Hence, the sixth term is \(12\). Option \(6\) is the fifth term, so it is a close but incorrect distractor. Exam tip: while finding the \(n\)th term, use \(a_n=a+(n-1)d\) carefully.

In an AP, the first term is \(a=28\) and the second term is \(a+d=19\). Therefore, \(d=19-28=-9\). Hence, the correct answer is \(-9\). Choosing \(9\) misses the negative sign of the difference. Exam tip: To find the common difference, subtract the previous term from the next term.

To find the common difference of an algebraic arithmetic progression, subtract one term from the next. Like terms cancel in the same way as ordinary numbers. If the result is negative, each successive term is smaller than the previous one by the absolute value of that negative number.

Using the first two terms, \\(d=(5m-4)-5m=-4\\). The same result appears again: \\((5m-8)-(5m-4)=-4\\), and \\((5m-12)-(5m-8)=-4\\). Thus the expression has a constant difference, so it is an arithmetic progression with common difference \\(-4\\). The variable part \\(5m\\) cancels; it is not the answer. Therefore choice B is correct.

The governing concept is the common difference of an arithmetic progression. For option A, subtract consecutive terms: 6−2=4, 10−6=4, and 14−10=4. Thus its first three consecutive differences are exactly (4,4,4), and it is an arithmetic progression with common difference 4. In option B the differences are 4, 5, and 6, so they change. In option C each difference is −2, not +4. In option D the differences are 3, 5, and 7, which also are not constant. Hence only option A satisfies the stated condition. The repeated wording in the supplied option has been normalized, without changing the mathematical choices or answer.

In an arithmetic progression, the common difference d is found by subtracting one term from the next consecutive term. Using the first two terms gives d = 29 - 24 = 5. The other pairs confirm the same result: 34 - 29 = 5 and 39 - 34 = 5. Therefore the sequence increases by 5 at every step, and option B is correct. Option A is one less than the actual difference, option C does not arise from any consecutive subtraction in the displayed sequence, and option D is the difference between neither adjacent pair. The calculation must use consecutive terms rather than the first and last terms.

The difference from (23) to (30) is (7), so (k=23-7=16). For a missing term use the same difference backward.

In an AP, \(a\) is the first term, so \(a=-8\). The common difference is the difference between consecutive terms: \(d=-3-(-8)=5\). Therefore, \((a,d)=(-8,5)\). The option \((-8,-5)\) results from an incorrect sign while subtracting a negative number. Exam tip: write \(d=a_2-a_1\) and use brackets around negative terms.

An arithmetic progression is defined by a constant difference between every pair of consecutive terms. For this sequence, the differences are 7 − 3 = 4, 13 − 7 = 6, and 21 − 13 = 8. Since 4, 6, and 8 are not equal, the sequence does not satisfy the defining condition. Therefore option C is correct. The first term may be any number, and positive terms can certainly form an arithmetic progression, so A and B are irrelevant. An infinite progression does not need a final term, so D is also incorrect. The terms increase, but mere increase is not enough; the amount of increase must remain constant.

In an arithmetic progression, the first term is \\(a\\), and each later term is obtained by repeatedly adding the common difference \\(d\\). Thus the second term is \\(a+d\\), while the third term is \\(a+2d\\). A negative common difference means that the repeated additions make the terms smaller.

Here \\(a=15\\) and \\(d=-4\\). The first term is 15, the second is \\(15+(-4)=11\\), and the third is \\(11+(-4)=7\\). Equivalently, the third-term formula gives \\(a_3=a+2d=15+2(-4)=15-8=7\\). Therefore choice B is correct. The value 11 is the second term, not the requested third term.

To find the common difference, subtract the first term from the second term: \((4x+3)-4x=3\). Checking the next pair, \((4x+6)-(4x+3)=3\) as well. Therefore, the common difference is 3. The number 6 is part of the third term, not the common difference. Exam tip: In an AP, the difference between every pair of consecutive terms must be the same.

The defining rule of an arithmetic progression is that consecutive differences are equal. From the known terms, 25 − 18 = 7 and 32 − 25 = 7, so the common difference is 7. Since p comes immediately before 18, it must satisfy 18 − p = 7. Solving gives p = 18 − 7 = 11, so option C is correct. Substituting this value produces 11, 18, 25, 32, …, whose consecutive differences are all 7. If p were 9, 10, or 12, the difference from p to 18 would be 9, 8, or 6 respectively, and the sequence would no longer have one constant common difference.

In an arithmetic progression, the common difference is \(d=a_2-a_1\). Here, \(d=51-56=-5\), so each successive term is 5 less than the preceding term. \(5\) is only the magnitude of the decrease; the common difference must be negative. Exam tip: always subtract the first of two consecutive terms from the second.

For three consecutive terms of an arithmetic progression, the middle term is the average of the first and third terms. Thus, \(m=\frac{8+20}{2}=14\). Therefore, 14 is correct. If 12 or 15 is chosen, the common differences on the two sides will not be equal. Exam tip: for consecutive AP terms \(a,b,c\), use \(2b=a+c\).

In option A, the differences between consecutive terms are \(53-60=-7\), \(46-53=-7\), and \(39-46=-7\). Since every difference is the same, it is an arithmetic progression with \(d=-7\). Option B has common difference \(+7\), while options C and D do not have equal consecutive differences. Exam tip: To identify an AP, check the difference between every pair of consecutive terms.

(\frac{7}{6}-\frac{5}{6}=\frac{1}{3}) and (\frac{3}{2}-\frac{7}{6}=\frac{1}{3}). For fractions subtract using common denominators.

In an arithmetic progression, each new term is obtained by adding the common difference d to the previous term. Here, a = -9 and d = 6: -9, -9+6 = -3, -3+6 = 3, and 3+6 = 9. Therefore, the first four terms are (-9, -3, 3, 9). Option D subtracts 6 each time, so its common difference is -6. Exam tip: check the difference between consecutive terms to verify an AP.

The common difference is found by subtracting a term from the next term. Here, \(1.9-1.4=0.5\) and \(2.4-1.9=0.5\). Since the difference between consecutive terms is the same, the common difference is \(0.5\). Option \(0.4\) is a distractor because it is not the difference between consecutive terms. Exam tip: Check at least two consecutive differences to confirm an AP.

In an arithmetic progression, the common difference \(d\) equals the later term minus the earlier term. Therefore, \(d=35-27=8\). \(-8\) would result only if the terms were in the order \(35, 27\). Exam tip: For two consecutive terms, subtract the first term from the second term.

The governing rule of an arithmetic progression is that each new term is obtained by adding the same common difference to the preceding term. Here, 19 − 12 = 7, 26 − 19 = 7, and 33 − 26 = 7, so the common difference is consistently 7. To find the term after 33, add this difference once more: 33 + 7 = 40. Thus option C is correct. Option A would add 5, option B would add 6, and option D would add 8. Each of those choices breaks the established pattern. The ellipsis means that the same rule continues beyond the displayed terms; it does not signal a change in the operation after 33.

For consecutive terms of an arithmetic progression, the difference between adjacent terms is constant. From the first two terms, the common difference is \(d=11-5=6\). Add this difference to the second term to obtain the third term: \(n=11+6=17\). Adding 6 once more gives \(17+6=23\), matching the fourth term and confirming the calculation.

Therefore \(n=17\), so option C is correct. The sequence is \(5,11,17,23\), and each term increases by 6. Values such as 15 or 16 would make the differences unequal, while 18 would produce a third-to-fourth difference of only 5. Hence only 17 satisfies the arithmetic-progression condition.