Introduction to APs and common difference.

The common difference is (23-18=5), and the same difference continues. In exams always check the difference of consecutive terms.

In an AP, \(a\) is the first term, so \(a=-2\). The common difference \(d\) is the difference between consecutive terms: \(d=3-(-2)=5\). Therefore, \(a=-2,\ d=5\) is correct. Option B incorrectly treats the second term, 3, as the first term. Exam tip: take extra care with signs when subtracting a negative number.

In an arithmetic progression, the difference between every pair of consecutive terms is constant. In option C, \(9-5=4\), \(13-9=4\), and \(17-13=4\). Hence, its common difference is \(4\), so it is an AP. In option B, the differences are \(3,5,7\), which are not equal. Exam tip: To identify an AP, quickly check consecutive differences.

The direct answer is option D: consecutive differences are not equal. An arithmetic progression requires one fixed common difference between every two consecutive terms. For the sequence 2, 6, 12, 20, calculate carefully: \(6-2=4\), \(12-6=6\), and \(20-12=8\). The differences are 4, 6, and 8, so they change rather than remain constant. Therefore the sequence is not an AP. Option A is wrong because the first term being 2 has no bearing on whether the differences are equal; an AP may start with any number. Option B is wrong because increasing terms alone do not guarantee an AP. For example, 1, 2, 4, 8 is increasing but its differences are 1, 2, 4, not equal. Option C is wrong because being positive also does not decide the type of sequence; an AP can contain positive, negative, or zero terms. Option D states the exact defining failure: the consecutive differences are unequal. The terms actually follow another pattern, since they can be written as products of consecutive integers, but that is not needed to answer this question. The key test is always subtraction of neighboring terms. Memory cue: AP means equal differences, not merely increasing terms or positive terms.

In an AP, every consecutive difference must be equal. Here, 21-25=-4 and 17-21=-4, but 14-17=-3. Hence it is not an AP. Exam tip: check at least all adjacent differences, not merely whether terms increase or decrease.

In an arithmetic progression, the common difference is the amount added to one term to get the next term. Here, from \(3\) to \(3+r\) the increase is \((3+r)-3=r\). The same holds for the next step: \((3+2r)-(3+r)=r\). So the common difference is \(r\). Hence option B is correct.

In an arithmetic progression, the common difference is d = second term − first term. Here, d = 27 − 31 = -4. Each successive term is 4 less than the preceding term, so the common difference is negative. Although 4 is the magnitude of the decrease, it is not the common difference. Exam tip: always subtract an earlier term from the next consecutive term to find d.

The middle term is the average of (3) and (11), so (m=\frac{3+11}{2}=7). In three consecutive terms the middle term is the average.

In option B, the differences between consecutive terms are \(34-40=-6\), \(28-34=-6\), and \(22-28=-6\). Hence, it is an arithmetic progression with \(d=-6\). In option C, the differences are \(-6,-4,-6\), so they are not constant. Exam tip: To identify an AP, check the difference between every pair of consecutive terms.

(1-\frac{3}{4}=\frac{1}{4}) and (\frac{5}{4}-1=\frac{1}{4}). For fractions subtract using common denominators.

In an arithmetic progression, the common difference d is added to each term to obtain the next term. Here, a = -5 and d = 4: -5, -5 + 4 = -1, -1 + 4 = 3, and 3 + 4 = 7. Therefore, the first four terms are (-5, -1, 3, 7). In option A, the common difference is 3, not 4. Exam tip: Always check the difference between consecutive terms.

To find the common difference, subtract a term from the term immediately after it: \(0.5-0.2=0.3\). Checking further, \(0.8-0.5=0.3\), so the common difference is 0.3. The number 0.2 is the first term, not the common difference. Exam tip: find the difference between consecutive terms and verify it with another pair.

The defining rule for an arithmetic progression is d = a next term − a preceding term. Since the consecutive terms are 12 followed by 17, calculate d = 17 − 12 = 5. Thus option C is correct. The order matters: subtracting the earlier term from the later term gives the forward common difference. Option A is the sum 12 + 17, not the difference; option B is merely the first supplied term; and option D is the second supplied term. Because the terms are consecutive, no formula for the sum or nth term is needed. The same difference would continue in the progression, so the next term after 17 would be 22.

In an arithmetic progression, the same number is added or subtracted to get each next term. That fixed number is called the common difference. In the given sequence, the terms increase regularly, so we can find the next term by adding the difference to the last displayed term. This makes option B the correct answer.

The consecutive differences are \(13-8=5\), \(18-13=5\), and \(23-18=5\). Therefore the common difference is 5. Adding it to the last term gives \(23+5=28\). Hence the next term is 28. The other choices do not continue the established pattern because they add 3, 7, or 8 instead of 5.

In an arithmetic progression, the common difference is the same between every pair of consecutive terms. The first two terms here are 6 and 10, so \(d=10-6=4\). The third term is obtained by adding this difference to the second term: \(n=10+4=14\). The fourth term then becomes \(14+4=18\), which confirms that the sequence is consistent.

Therefore \(n=14\), so option D is correct. Another way to see this is that the middle term of four consecutive terms is equally spaced: the terms move from 6 to 10, then to 14, and then to 18. The values 12, 13, and 15 do not preserve the same difference throughout the progression.

The third term is (36) and the fourth is (33), so (33-36=-3). While finding the difference subtract the previous term from the next term.

In an arithmetic progression, the second term is obtained by adding the common difference to the first term: \(a+d=2.5+1.5=4.0\). Therefore, 4.0 is correct. The value 3.5 would result from adding only 1 to 2.5, not the given common difference of 1.5. Exam tip: For the second term of an AP, use \(a+d\) directly.

In the sequence (11,11,11,11,\ldots), every term is equal to the preceding term. Therefore, the common difference is \(d=11-11=0\). Option B has common difference \(-5\), not zero. Exam tip: To find the common difference of an AP, subtract the first term from the second term.

The first term is x and the second term is x+7. Thus, second term − first term = (x+7)-x=7. Therefore, the common difference is 7. Note that x+7 is the second term, not the difference. Exam tip: To find the common difference of an AP, subtract a term from the term immediately after it.

In this AP, the first term is \(a=16\) and the common difference is \(d=20-16=4\). Therefore, \(a+d=16+4=20\), which is the second term of the AP. \(24\) is the third term, so it is not correct. Exam tip: \(a+d\) is always the second term of an AP.

In an arithmetic progression, the difference between consecutive terms is constant. Here, the common difference is \(d=18-21=-3\). Therefore, the third term is \(q=18+(-3)=15\). Checking: \(15-3=12\), so \(21,18,15,12\) is correct. If \(14\) were used, the consecutive differences would not be equal. Exam tip: Find \(d\) from the first two terms and apply the same difference to obtain the next term.

Every consecutive difference is (5), so it is an arithmetic progression. While choosing the statement check both type and (d).

In an arithmetic progression, the common difference is d = a₂ − a₁. Thus, d = 8 − 13 = -5. Therefore, -5 is correct. Choosing 5 would reverse the order of subtraction. In exams, subtract the first term from the second term to find d.

In the general form (a,a+d,a+2d,\ldots), (a+2d) is the third term. Remembering the general form makes term identification easier.

This is an AP since 11−15=−4, 7−11=−4, and 3−7=−4. A negative common difference makes terms decrease but does not stop it from being an AP. Exam tip: compare consecutive differences.