Introduction to APs and common difference.

The second term of an arithmetic progression is given by \(a_2=a+d\). Thus, \(a_2=9+(-2)=7\). The value 9 is the first term, while 11 results from incorrectly adding 2 instead of adding the negative common difference. In exams, pay careful attention to the sign of the common difference.

The common difference of an arithmetic progression is the difference between two consecutive terms. Here, 90 − 100 = −10 and 80 − 90 = −10, so the common difference is −10. Since the sequence is decreasing, its common difference is negative, not simply 10. Exam tip: subtract the preceding term from the succeeding term.

The common difference of an arithmetic progression is the difference between consecutive terms. In option B, \(16-20=-4\), \(12-16=-4\), and \(8-12=-4\), so its common difference is \(-4\). In options A and D, 4 is added each time, while the differences in option C are not constant. Exam tip: subtract each term from the next one and check whether the difference remains the same.

The governing concept is that an arithmetic progression (AP) has the same difference between every pair of consecutive terms. Subtract the first term from the second term: 3 − 0 = 3. Checking the next pair gives 6 − 3 = 3, and the following pair gives 9 − 6 = 3, confirming that the difference is constant. Therefore, the common difference d is 3, so option B is correct. Option A is the first term, not the difference; option C is a later term; and option D is another term in the sequence. The fact that the first term is zero does not change the common difference.

For an arithmetic progression, a denotes the first term and d denotes the common difference between consecutive terms. The first listed term is therefore a = 11. To find d, subtract any term from the term immediately after it: 15 − 11 = 4. The other pairs confirm the same result: 19 − 15 = 4 and 23 − 19 = 4. Since the question asks for d and a in that order, the answer must be d = 4, a = 11, which is option B. Option A reverses the meanings of a and d; option C treats a later term as the difference; and option D incorrectly takes 15 as the first term.

In an arithmetic progression, the difference between consecutive terms is constant and is denoted by \(d\). When \(d=0\), no change occurs from one term to the next, so the progression is \(a, a, a, a, \ldots\). Hence, all terms are equal. The terms need not all be negative; that depends on the first term. Exam tip: Recognise an AP with \(d=0\) as a constant progression.

In option B, the differences between consecutive terms are equal: \(8-3=5\), \(13-8=5\), and \(18-13=5\). Thus, its common difference is 5, so it is an arithmetic progression. In option A and option D, the terms are multiplied by 2, while option C has differences 3, 5, and 7; therefore, none of them is an arithmetic progression. For exams, calculate the differences between consecutive terms to identify an AP quickly.

The common difference of an arithmetic progression is found by subtracting a term from the term immediately before it. Here,
(-6)-(-10)=-6+10=4. The next differences are also
(-2)-(-6)=4 and 2-(-2)=4, so the common difference is 4. Exam tip: when subtracting a negative number, its sign changes to plus.

The common difference of an arithmetic progression is the difference between two consecutive terms. Here, the next term is 23 and the previous term is 18, so the common difference is 23 − 18 = 5. Therefore, option A is correct. Exam tip: use common difference = next term − previous term.

The differences between consecutive terms are 8−4=4, 13−8=5, and 19−13=6. In an arithmetic progression, the difference between every pair of consecutive terms must be constant. Since these differences are 4, 5, and 6, the sequence is not an arithmetic progression. Merely having increasing or positive terms is not sufficient. Exam tip: calculate consecutive differences to test whether a sequence is an AP.

To find the common difference, subtract each term from the term immediately after it. Here, \(4-1=3\), \(7-4=3\), and \(10-7=3\). Therefore, each term is 3 greater than the previous term, so the common difference is 3. Exam tip: In an AP, the difference between consecutive terms remains constant.

In an arithmetic progression, the first term is \(a=2\) and the common difference is \(d=5\). Add 5 successively: \(2\), \(2+5=7\), and \(7+5=12\). Therefore, the correct sequence is (2, 7, 12). Option A does not maintain a common difference of 5. Exam tip: start with the first term and add the common difference to obtain each next term.

An arithmetic progression is determined by two pieces of information: its first term \(a\) and its common difference \(d\). Starting with \(a=9\), add \(d=2\) repeatedly. This gives the pattern \(9, 9+2, 9+2+2, 9+2+2+2,\ldots\). Equivalently, its general term is \(a_n=a+(n-1)d\), so here \(a_n=9+2(n-1)\).

The first listed sequence is \(9,11,13,15,\ldots\). It begins with 9, and each consecutive difference is 2: \(11-9=2\), \(13-11=2\), and \(15-13=2\). Therefore choice A matches both requirements. Option C begins with 9 but has common difference 9, while option B begins with 2 and option D begins with 11, so none of those satisfies both conditions.

The difference between consecutive terms is constant: \(35-40=-5\), \(30-35=-5\), and \(25-30=-5\). Hence, this is an AP with common difference \(-5\). Therefore, the terms after 25 are \(25-5=20\) and \(20-5=15\). In option (22, 19), the difference is \(-3\), so it does not continue the AP. Exam tip: Find the difference between consecutive terms first, then apply the same difference to get the next terms.

The explicitly listed terms are 18, 21, 24 and 27. The common difference is \(21-18=3\), so the next term would be \(27+3=30\). Therefore, 30 is a term of the AP but is not among the terms given in the question. Exam tip: distinguish the terms already listed from the terms that can be generated next.

The common difference of an arithmetic progression is the difference between consecutive terms. Here, \(d=5-5=0\), and the difference between every pair of consecutive terms is also zero. Therefore, all the terms are equal, so it is an arithmetic progression with constant terms. It is neither increasing nor decreasing. Exam tip: an AP with \(d=0\) is called a constant AP.

In an arithmetic progression, the common difference is the difference between two consecutive terms. Here, \(14-12=2\) and \(16-14=2\), so the common difference is 2. Option 4 is incorrect because it is the difference between the first and third terms, not consecutive terms. Exam tip: subtract each term from the term immediately after it to find the common difference.

Find the differences between consecutive terms: \(45-50=-5\) and \(40-45=-5\). Thus, each term is 5 less than the preceding term, and the common difference is \(-5\). Since the question asks how much less, the answer is 5 rather than \(-5\). Exam tip: “how much less” gives the positive magnitude of the decrease, whereas the common difference is negative.

The defining rule of an arithmetic progression is that consecutive terms differ by one fixed number. Since all the displayed fractions have the same denominator, subtract their numerators and retain the denominator. Thus, the difference between the first two terms is 5/4 − 3/4 = 2/4 = 1/2. The next differences agree: 7/4 − 5/4 = 2/4 = 1/2 and 9/4 − 7/4 = 1/2. Hence the common difference is 1/2, making option B correct. Option A results from subtracting the numerators incorrectly as one unit; option C is the first term; and option D is not the difference between consecutive terms.

In an arithmetic progression, the common difference is found by subtracting the first term from the second term: d = a₂ − a₁ = 10 − 16 = −6. Therefore, option C is correct. Option B gives only the magnitude, 6, and misses the negative sign. Exam tip: when the next term is smaller than the previous term, the common difference is negative.