Introduction to APs and common difference.

In an arithmetic progression, the common difference is the difference between two consecutive terms. Here, the second term is 25 and the first term is 18, so the common difference is 25 - 18 = 7. The value 43 is the sum of the terms, not their difference. Exam tip: use next term − previous term.

The common difference is found by subtracting a term from the term immediately after it. Here, \(2-\frac{3}{2}=\frac{1}{2}\) and \(\frac{5}{2}-2=\frac{1}{2}\). Therefore, the common difference is \(\frac{1}{2}\). The value \(1\) is not the increase between consecutive terms. Exam tip: In an AP, the difference between every pair of consecutive terms must be the same.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_3=20+(3-1)(-3)=20-6=14\). The value 17 is the second term, not the third. Exam tip: when \(d\) is negative, each successive term decreases.

The first term of the AP is 9 and the common difference is -2. Therefore, 2 is subtracted to get each next term: 9, 7, 5, 3, \ldots. Hence, option B is correct. Option D also has common difference -2, but its first term is 7, not 9. Exam tip: check the first term first, then find the difference between consecutive terms.

The common difference is found by subtracting a term from the term immediately after it. Here, \(1.8-1.2=0.6\) and \(2.4-1.8=0.6\). Therefore, the equal difference between consecutive terms is \(0.6\). Although \(0.4\) is a close distractor, it is not the difference between any two consecutive given terms. Exam tip: Verify the common difference using at least two consecutive pairs.

Each term in the given sequence is 6 greater than the previous term, so the common difference is 6. The fourth term is 20; therefore, the fifth term is 20 + 6 = 26. Choosing 24 is incorrect because it adds 4 to 20, whereas the required difference is 6. Exam tip: In an AP, find the difference between consecutive terms to obtain the next term quickly.

In an arithmetic progression, the common difference is the difference between consecutive terms. Here, \(4-4=0\), so the difference between every pair of consecutive terms is \(0\), and hence \(d=0\). If \(d=4\), each next term would increase by 4. Exam tip: Find the common difference using second term − first term.

The consecutive differences are equal, so the sequence is an arithmetic progression. In exams, first check equality of differences.

The difference between consecutive terms is constant: \(92-100=-8\), \(84-92=-8\), and \(76-84=-8\). Hence, this is an arithmetic progression with common difference \(d=-8\). Therefore, the terms after 76 are \(76-8=68\) and \(68-8=60\). In option A, the difference is \(-6\), so it does not follow the rule of the sequence. Exam tip: To find subsequent terms, first check the difference between consecutive terms.

In option A, the difference between consecutive terms is constant: \(12-5=7\), \(19-12=7\), and \(26-19=7\). Therefore, its common difference is \(d=7\). Option B has common difference \(5\), whereas option C has common difference \(-7\). Exam tip: Subtract the first term from the second term, then verify with the next pair of terms.

In an arithmetic progression, the common difference is the difference between consecutive terms. Here, \((x+4)-x=4\) and \((x+8)-(x+4)=4\), so the common difference is \(4\). \(8\) is the difference between the first and third terms, not between consecutive terms. Exam tip: subtract consecutive terms to find the common difference.

In an arithmetic progression, consecutive differences are equal. Here the differences change, so it is not an arithmetic progression.

The saving on the first day is the first term, so \(a=50\). The saving increases by \(15\) rupees each day, so the common difference is \(d=15\). Thus the terms are \(50, 50+15=65, 65+15=80, 80+15=95\), giving \((50,65,80,95,\ldots)\). Option D has a difference of \(15\), but it starts with \(65\), not \(50\). Exam tip: In an AP, always check both the first term and the difference between consecutive terms.

The first term of the given AP is \(a=3\). Its common difference is \(d=9-3=6\). Hence, \(a+d=3+6=9\). Note that \(6\) is only the common difference, not the required sum. Exam tip: Find \(d\) by subtracting a term from the term immediately after it.

In an arithmetic progression, the common difference is \(d=a_2-a_1\). Thus, \(d=29-35=-6\). Therefore, \(-6\) is correct. Choosing \(6\) misses the negative sign. Exam tip: always subtract the preceding term from the succeeding term.

The key idea is to identify the position of a term by counting from the left. In any listed sequence, the first displayed value is the first term, the next value is the second term, and so on. The fractions do not need to be converted into decimals because the question asks only for the term at a stated position.

Count the terms carefully: the first term is \\(5/4\\), the second is \\(7/4\\), the third is \\(9/4\\), and the fourth is \\(11/4\\). Thus the fourth term is \\(11/4\\), which is option D. Although the sequence has a common difference of \\(1/2\\), calculating it is unnecessary here because the fourth term is already written explicitly. Options A, B, and C are the first three terms, not the fourth.

An arithmetic progression has one fixed common difference throughout the sequence. For a required difference of \\(d=-5\\), every next term must be exactly 5 less than the preceding term. The minus sign is important: the sequence must decrease by 5 each time, not increase by 5 or decrease by varying amounts.

For choice B, the consecutive differences are \\(35-40=-5\\), \\(30-35=-5\\), and \\(25-30=-5\\). Since the same value continues, it is an arithmetic progression with common difference \\(-5\\). Choice A has difference \\(+5\\), choice C changes by \\(-5,-6,-7\\), and choice D has differences \\(-5,-10,-5\\). Therefore only choice B satisfies the condition.

In an arithmetic progression, each term is obtained by adding the common difference d to the preceding term. When d>0, every next term is greater than the previous one, so the terms increase. They remain equal only when d=0, and they decrease when d<0. Exam tip: Use the sign of d to identify the trend of an AP quickly.

The first term of the sequence is \(a=0\). The common difference is the difference between consecutive terms: \(5-0=5\), so \(d=5\). Therefore, in the order \(d\) and \(a\), the answer is \(d=5,\ a=0\). \(d=10\) is incorrect because it is the difference between the first and third terms, not between consecutive terms. Exam tip: always use \(a=a_1\) and \(d=a_2-a_1\).

In an AP, the difference between every pair of consecutive terms must be the same. Here, 5−2=3, 9−5=4, and 12−9=3, so the sequence is not an AP. Exam tip: check consecutive differences first.