Introduction to APs and common difference.

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.

For an AP, \(d\) is the difference between any two consecutive terms. \( d=\frac{2}{3}-\frac{1}{3}=\frac{1}{3} \). Also, \( 1-\frac{2}{3}=\frac{1}{3} \), so \( d=\frac{1}{3} \).

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.

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.

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.

For an A.P., the common difference \(d\) is the constant value of \(\text{next term} - \text{previous term}\). Using the first two terms: \[d=(3m-6)-(3m)=-6.\] The \(3m\) cancels, leaving a constant difference \(-6\). Therefore option B is correct.

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.

The \(n\)th term of an AP is \(a_n=a+(n-1)d\). So the third term is \(a_3=a+2d=15+2(-4)=15-8=7\).

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.

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 \(2b=a+c\).