Introduction to APs and common difference.

The governing concept is that an arithmetic progression has a constant difference between every pair of consecutive terms. Subtract successive terms to test the sequence: 7 - 3 = 4, 11 - 7 = 4, and 15 - 11 = 4. Since the same value is obtained each time, the sequence is an arithmetic progression and its common difference is d = 4. Therefore option C is correct. Option A and option B do not equal the change from one term to the next. Option D, although it appears as a term in another context, is itself the second term here and is not the fixed increment. Checking more than one pair is useful because a single subtraction alone would not establish that the difference remains constant throughout the sequence.

In an arithmetic progression, the common difference is the difference between consecutive terms. If the first two terms are \(a_1\) and \(a_2\), then \(d=a_2-a_1\). Option C reverses the subtraction and gives \(-d\). In exams, subtract the earlier term from the later term.

The common difference is found by subtracting a term from the next term: \(d=8-10=-2\). The other consecutive differences are also \(6-8=-2\) and \(4-6=-2\), so the progression is decreasing with a negative common difference. Option A has the wrong sign. Exam tip: use \(d=a_{n+1}-a_n\), not the reverse subtraction.

The direct answer is option C, \(5,10,15,20,\ldots\). An arithmetic progression has a constant difference between consecutive terms. For option C, \(10-5=5\), \(15-10=5\), and \(20-15=5\), so the common difference is 5. Option A doubles each time: differences are 2 and 4, so they are not constant. Option B contains square numbers; its differences are 3, 5, and 7, so it is not an AP. Option C has equal differences and is correct. Option D also doubles: differences are 3, 6, and 12, so it is not an AP. Do not confuse a constant ratio with a constant difference; doubling describes a geometric pattern, not an arithmetic one. Memory cue: subtract neighbouring terms; if every answer is the same, it is an AP.

The difference of consecutive terms is (6-6=0). A constant sequence is also an arithmetic progression with common difference (0).

The first term of an arithmetic progression is the number with which the sequence begins. In the given progression, the first term is 12, so option B is correct. The number 17 is the second term, while 5 is the common difference, not the first term. Exam tip: To identify the first term, look at the leftmost number in the progression.

This is an arithmetic progression because the difference between consecutive terms is constant: \(9-4=5\) and \(14-9=5\). Hence, with common difference \(5\), the missing term is \(14+5=19\). The values 18 and 20 do not maintain the same common difference. Exam tip: first check the difference between consecutive known terms.

An arithmetic progression is defined by a constant common difference between consecutive terms. Test each option systematically. In A, the differences are 11-7 = 4, 15-11 = 4, and 19-15 = 4, so it is an AP. In B, the differences are -3, -3, and -3, so it is also an AP, even though it decreases. In C, the differences are 2, 2, and 2, so it qualifies. In D, the differences are 5-2 = 3, 9-5 = 4, and 14-9 = 5; they change from one step to the next. Hence D is not an arithmetic progression, and option D is correct.

In an arithmetic progression, the second term is the sum of the first term and the common difference: second term = a + d = 5 + 3 = 8. Therefore, option A is correct. Option D is only the first term, not the second term. Exam tip: remember the formula a₂ = a + d for finding the second term.

The common difference is the difference between two consecutive terms. Here, \(d=a_2-a_1=1-(-3)=4\); the other consecutive differences are also \(5-1=4\) and \(9-5=4\). Therefore, the correct answer is 4. The value 5 is a term, not the difference between consecutive terms. Exam tip: subtract any two consecutive terms to verify the common difference.

The common difference of an arithmetic progression is found by subtracting a term from the term immediately before it. Here, \\(20-25=-5\\) and \\(15-20=-5\\), so the common difference is \\(-5\\). Option A has the wrong sign; a decreasing progression has a negative common difference. Exam tip: subtract consecutive terms and always check the sign.

The common difference of an arithmetic progression is defined as the difference between a term and the term immediately before it: d=a₂−a₁. Here every new term is obtained by adding 7, so a₂−a₁=7 and the same calculation holds for every consecutive pair. Therefore d=7, making option C correct. Zero would mean no change, and −7 would describe a decreasing sequence.

The common difference of an arithmetic progression is the fixed amount added to obtain each term from the preceding term. It is found by subtracting one term from the next, such as \(a_2-a_1\). The terms may be whole numbers, fractions, or decimals; the method remains exactly the same. Every pair of consecutive terms should give the same difference.

For this sequence, subtract the first term from the second: \(2.0-1.5=0.5\). The next checks agree: \(2.5-2.0=0.5\) and \(3.0-2.5=0.5\). Thus the constant difference is \(0.5\), so choice B is correct. The value \(0.25\) would be too small, while \(1.0\) and \(1.5\) do not equal the change between consecutive terms.

The defining property of an arithmetic progression is a fixed difference between consecutive terms. Subtract the first term from the second: d = 1 - 1/2 = 1/2. Verify it with the remaining displayed pairs: 3/2 - 1 = 1/2 and 2 - 3/2 = 1/2. Since every step adds one-half, the common difference is 1/2, so option A is correct. The numbers 1, 3/2, and 2 are terms of the progression, not the fixed increment. A common error is to confuse the next term with the difference or to subtract the numerators without considering the denominator. Writing 1 as 2/2 makes the first subtraction especially clear: 2/2 - 1/2 = 1/2.

In option B, the differences between consecutive terms are equal: \(7-4=3\), \(10-7=3\), and \(13-10=3\). Therefore, it is an arithmetic progression with common difference \(d=3\). Option C has common difference 4, while option D has common difference −3. In an exam, find the common difference by subtracting each term from the term immediately after it.

In an arithmetic progression, subtract one term from the next to obtain the constant difference. Using the first two terms, d=(a+2)−a=2. The next checks give (a+4)−(a+2)=2 and (a+6)−(a+4)=2, confirming that the difference is constant. Therefore option C is correct; a and 2a are expressions involving the first term, not the term-to-term increase.

The common difference is the difference between consecutive terms: \(27-30=-3\), \(24-27=-3\), and \(21-24=-3\). Therefore, the next term is \(21+(-3)=18\). Option 19 is incorrect because it would give a difference of \(-2\), whereas the sequence decreases by 3 each time. Exam tip: identify the common difference first, then add it to the last given term.

Count the terms from left to right: the first term is 2, the second is 5, and the third is 8. The common difference is 3, but for this question, identifying the third position directly is sufficient. Therefore, the correct answer is 8; 5 is the second term and 11 is the fourth term. In an exam, do not confuse the term number with its value.

An arithmetic progression is a sequence in which the difference between consecutive terms remains constant. For example, in 3, 7, 11, 15, each subtraction gives 4, so d=4. The terms need not have equal squares or be zero. A rule in which each term is double the previous one describes a geometric progression, not an arithmetic progression. Hence option B states the defining property.

In an arithmetic progression, the difference between every pair of consecutive terms must be constant. Here, the differences are 6−3=3, 12−6=6, and 24−12=12. Since these are not equal, the sequence is not an arithmetic progression. In an exam, first check consecutive differences; being positive or integral does not make a sequence an AP.

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 the common difference d of an arithmetic progression is the constant amount added to one term to obtain the next. For this sequence, subtract consecutive terms: 3 - 0 = 3, 6 - 3 = 3, and 9 - 6 = 3. The repeated result confirms that the progression has d = 3. Therefore option B is correct. Option A is the first term, not the change between terms. Options C and D are later terms in the sequence and do not represent the fixed increment. The fact that the sequence begins with zero does not make the common difference zero; it only identifies the first term. A difference of zero would produce a constant sequence such as 0, 0, 0, which is not the sequence given here.

For an arithmetic progression, a denotes the first term and d denotes the constant difference between consecutive terms. The first term listed is 11, so a = 11. Calculate the common difference using adjacent terms: d = 15 - 11 = 4. The result is confirmed by 19 - 15 = 4 and 23 - 19 = 4. Because the question asks for d and a in that order, the required pair is d = 4, a = 11, which is option B. Option A reverses the meanings of the symbols. Option C incorrectly treats a later term as d, while option D uses the second term as a even though a must be the first term. Careful attention to the requested order is essential.