Introduction to APs and common difference.

Number the terms from left to right starting with position 1: a₁=14, a₂=20, a₃=26, and a₄=32. Therefore 20 is the second term, so option B is correct. The common difference is 6, but it is not necessary to calculate it for this question. The first term is 14, and 26 and 32 occupy the third and fourth positions respectively.

The first term of the AP is \(a=17\). The common difference is found by subtracting the previous term from the next term: \(d=13-17=-4\). Therefore, \(a\times d=17\times(-4)=-68\), so option B is correct. \(-4\) is only the common difference, not the required product. Exam tip: Find the common difference using next term − previous term.

For a sequence to be an arithmetic progression, the difference between every pair of consecutive terms must be one constant number. It is not enough for the terms to increase, decrease, or have one matching difference. The test must be applied to all consecutive differences that are provided.

Here the consecutive differences are listed as 2, 4, and 6. Because these numbers are not equal, there is no single constant common difference. Therefore the sequence is not an arithmetic progression, which makes choice B correct. Choice C is not correct because 2 is only the first difference, not the common difference of the whole sequence. Equal terms would instead give difference 0, and that is not the situation here.

The relevant concept is the common difference of an arithmetic progression. Compute each requested difference directly: the difference of the first two terms is 12 - 5 = 7, while the difference of the third and fourth terms is 26 - 19 = 7. Since both results are equal, the sequence is consistent with an arithmetic progression having d = 7. Option A reports both calculated differences. Option B incorrectly uses 5, which is the first term rather than a difference; option C lists terms instead of subtracting them; and option D doubles the second difference without justification. Hence option A is correct.

In an arithmetic progression, the same common difference d is added successively. Here, a = 3 and d = 0, so the second term is 3 + 0 = 3 and the third term is also 3 + 0 = 3. Therefore, the first three terms are (3, 3, 3). Option D has common difference 3, not 0. Exam tip: When d = 0, every term of an AP equals its first term.

In an AP, the first term is \(a=15\) and the common difference is \(d=-4\). Therefore, subtract 4 to get each next term: \(15,\;15-4=11,\;11-4=7,\;7-4=3\). Hence, the first four terms are \((15, 11, 7, 3)\). Option D starts from 11, so it omits the first term. Exam tip: a negative \(d\) means that the terms decrease.

To find the common difference, subtract a term from the term immediately after it. Here, \((2k+3)-2k=3\) and \((2k+6)-(2k+3)=3\). Thus, the difference between every pair of consecutive terms is \(3\). \(2k+3\) is the second term, not the common difference. Exam tip: In an AP, check the differences of at least two consecutive pairs of terms to confirm that they are equal.

The second term of an AP is \(a_2=a_1+d\). Here, \(a_1=10\) and \(d=-1.5\), so \(a_2=10+(-1.5)=8.5\). The value 9.5 would result if the common difference were \(-0.5\), so it is not correct. Exam tip: adding a negative common difference decreases the term.

The common difference of an arithmetic progression is found by subtracting one term from the following term. For the given sequence, use the first two terms: \(d=\frac{2}{3}-\frac{1}{3}\). The denominators are already equal, so subtracting the numerators gives \(d=\frac{1}{3}\). Checking the next pair gives \(1-\frac{2}{3}=\frac{1}{3}\), and the following pair gives \(\frac{4}{3}-1=\frac{1}{3}\).

Thus the common difference is \(\frac{1}{3}\), so option A is correct. The value \(\frac{2}{3}\) is the second term, not the difference. Similarly, 1 and \(\frac{4}{3}\) are later terms. Equal differences confirm that the listed sequence is an arithmetic progression.

The difference between consecutive terms is the same: \(22-18=4\), \(26-22=4\), and \(30-26=4\). Hence, this is an arithmetic progression with common difference \(4\). Although \(2\) is a possible number, it is not the difference between any two consecutive terms here. Exam tip: To find the common difference, subtract the first term from the second term.

The difference between consecutive terms is constant: \(57-64=-7\). Hence, this is an AP with common difference \(d=-7\). Since the fourth term is 43, the fifth term is \(43-7=36\). Option 37 would result from incorrectly using a common difference of \(-6\). Exam tip: To find the next term of an AP, add the common difference to the preceding term.

Given \(a=4\) and \(a+d=13\). Thus, \(4+d=13\), so \(d=13-4=9\). In an AP, \(d\) is the common difference. \(13\) is the second term here, not the common difference. Exam tip: when \(a+d\) is given, subtract \(a\) from it to find \(d\).

In option B, the difference between consecutive terms is constant: \(0.5-0.25=0.25\), \(0.75-0.5=0.25\), and \(1.0-0.75=0.25\). Therefore, it is an AP with \(d=0.25\). Option A has common difference \(0.5\), while option C does not have a constant difference. Exam tip: To identify an AP, check the difference between every pair of consecutive terms.

The common difference of an AP is found by subtracting a term from the term immediately after it. Here, \(7.5-7=0.5\) and \(8-7.5=0.5\), so the common difference is \(0.5\). The value \(1\) is not the difference between consecutive terms. In exams, verify the difference using any two consecutive terms.

The second term of the sequence is 4 and the third term is 7. Hence, the difference between the third and second terms is \(7-4=3\). Therefore, 3 is correct. The number 4 is the second term, not the difference. Exam tip: To find the difference between consecutive terms, subtract the earlier term from the later term.

The common difference of an AP is found by subtracting a term from the term immediately after it. Here, \((a+1)-(a-2)=a+1-a+2=3\). Also, \((a+4)-(a+1)=3\), so the common difference is \(3\). \(a+3\) could represent a term, not the difference. Exam tip: When subtracting algebraic expressions, use brackets carefully to handle signs correctly.

In the sequence (2,3,5,8,\ldots), the differences between successive terms are \(3-2=1\), \(5-3=2\), and \(8-5=3\). Since these differences are not equal, it is not an arithmetic progression. In contrast, the common differences in A, B, and C are \(5\), \(-4\), and \(0\), respectively. Exam tip: subtract consecutive terms; a sequence is an AP only when this difference remains constant.

In an arithmetic progression, the common difference is d = a₂ − a₁. Here, a₁ = 11 and a₂ = 18, so d = 18 − 11 = 7. Therefore, the correct answer is 7. The value 6 is incorrect because it is not the difference between 11 and 18. Exam tip: To find the common difference, subtract an earlier term from the next term.

The position of a term is counted from the beginning of the sequence. Here, 25 is the first term, 21 follows it, so 21 is the second term. The common difference is -4, confirming that the displayed order is 25, 21, 17, 13. Therefore, option B is correct; options A, C and D identify the positions of 25, 17 and 13 respectively.

In an AP, the first term is a, and the common difference d is added to obtain each next term. Here the first three terms are 2.5, 2.5 + 1.5 = 4.0, and 4.0 + 1.5 = 5.5. Therefore, option A is correct. In option C, the first term is correct, but its common difference is 2.5, not 1.5. Exam tip: For the first three terms, use a, a + d, and a + 2d.