Introduction to APs and common difference.

In an arithmetic progression formed by repeated weekly savings increases, the common difference is the fixed amount added from one week to the next. Subtract the first two listed savings amounts: 150 − 100 = 50 rupees. The same result appears in the following steps: 200 − 150 = 50 and 250 − 200 = 50. Thus the sequence has common difference d = 50 rupees, so option A is correct. The amount 100 is the initial saving, not the weekly increase. The values 150 and 250 are later terms of the progression, so they cannot represent its common difference. The real-life wording does not change the AP rule.

The common difference is the difference between two consecutive terms. Here, \(45-40=5\), \(50-45=5\), and \(55-50=5\). Therefore, the common difference of the ticket prices is 5. Option 10 is incorrect because it is not the difference between consecutive terms. Exam tip: To find the common difference in an AP, subtract the first term from the second term.

The second term is 10 and the third term is 14. Therefore, their difference is 14 − 10 = 4. This is an arithmetic progression, so the difference between consecutive terms is constant. Exam tip: subtract the earlier term from the later term when finding the difference between consecutive terms.

In an arithmetic progression, the difference between every pair of consecutive terms is constant. In option B, 14−18=−4, 10−14=−4, and 6−10=−4, so its common difference is d=−4. Option A has a difference of +4, while the differences in C and D are not constant. An easy exam check is to subtract each term from the term immediately before it.

The common difference is found by subtracting one term from the next consecutive term. Here,
\((x+5)-x=5\), so the common difference is 5. Option D, \(x+5\), is a term of the sequence, not the common difference. Exam tip: In an AP, the difference between every pair of consecutive terms remains constant.

The common difference of an arithmetic progression is the difference between two consecutive terms. Here, \(d=(2p+3)-2p=3\), and the next difference is also \((2p+6)-(2p+3)=3\). Therefore, the correct answer is 3; 6 is the total increase from the first to the third term, not the common difference. Exam tip: subtract the first term from the second term to find the common difference.

In (2,4,7,11,\ldots), the differences are (2,3,4). To identify an arithmetic progression all differences must be equal.

First identify the common difference by subtracting consecutive terms. We have −7 − (−10) = 3, −4 − (−7) = 3, and −1 − (−4) = 3. Therefore this is an arithmetic progression with d = 3. To obtain the next term, add the same difference to the last displayed term: −1 + 3 = 2. Hence option A is correct. Option B would result from adding 2 instead of the established difference; option C moves in the wrong direction and ignores that the sequence is increasing; and option D adds 4. Careful handling of the negative sign is important: subtracting a negative number is equivalent to adding its positive value.

In an arithmetic progression, the second term is found by adding the common difference to the first term: second term = first term + d = 1 + 4 = 5. Therefore, 5 is correct. Option 4 is only the common difference, not the second term. Exam tip: In an AP, the difference between consecutive terms is always d.

To find the equal increase, subtract each term from the term immediately after it: \(30-25=5\) and \(35-30=5\). Therefore, the common difference of this sequence is \(d=5\). Option 10 is incorrect because it is not the difference between consecutive terms. Exam tip: In an AP, use \(d=a_{n+1}-a_n\).

Each successive term is 4 less than the preceding term: 12−16=−4, 8−12=−4, and 4−8=−4. Therefore, the magnitude of the equal decrease is 4. Note that the common difference of the AP is −4, whereas the decrease asked here is 4. Exam tip: subtract consecutive terms to verify the equal decrease.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Here, \(a=9\) and \(d=0\), so the second term is \(9+0=9\) and the third term is \(9+2\times0=9\). Therefore, the first three terms are \((9,9,9)\). Option B would represent a common difference of \(-9\), not 0. Exam tip: when the common difference is zero, every term of the AP equals the first term.

Each consecutive time has the same difference of (8), so it is an arithmetic progression. In word problems identify the equal difference.

The common difference is second term minus first term, so (11-7=4). In exams do not reverse the order of subtraction.

The common difference in an arithmetic progression is the change from one term to the next. It is calculated by subtracting an earlier term from the following term. If the sequence is truly arithmetic, this result stays the same for all consecutive pairs. This idea works equally well for increasing and decreasing sequences, and the difference may be positive, zero, or negative.

For the given sequence, subtract the first term from the second: \(8-4=4\). The other differences confirm it: \(12-8=4\) and \(16-12=4\). Hence the common difference is \(4\), making choice B correct. The number 8 is a term of the sequence, but it is not the common difference; this distinction is important. Values 2 and 12 also do not represent the change between consecutive terms.

The common difference of an arithmetic progression is the difference between two consecutive terms: second term − first term. Here, \(d=12-15=-3\), and the same value is obtained from \(9-12=-3\) and \(6-9=-3\). Therefore, option B, \(-3\), is correct. Since the progression is decreasing, its common difference is negative; writing only its magnitude, \(3\), would be incorrect. Exam tip: calculate \(a_2-a_1\) or subtract any term from the term immediately following it.

For an arithmetic progression, the common difference d is the difference between any term and the immediately preceding term: d = a₂ − a₁. Here the first two terms are both 9, so d = 9 − 9 = 0. The same calculation works for every consecutive pair: 9 − 9 = 0. A constant sequence is therefore a valid arithmetic progression whose common difference is zero; it does not need to increase or decrease. Option A confuses the term itself with the difference, option B would describe a sequence increasing by one, and option D would describe a decreasing sequence. Thus option C is correct.

The first term of an arithmetic progression is the number written at the beginning of the sequence. In 6, 13, 20, 27, \ldots, the first term is therefore 6. The number 7 is the common difference, while 13 is the second term. Exam tip: identify the first term by locating the first number in the progression.

This is an arithmetic progression with common difference \(11-5=6\). Therefore, the missing term is \(17+6=23\). Options 21 and 22 do not maintain a difference of 6 between consecutive terms. In an exam, find the common difference from two consecutive terms and add it to the preceding term.

In an arithmetic progression, the second term is the sum of the first term and the common difference: \(a_2=a+d\). Thus, \(a_2=7+4=11\), so 11 is correct. The options 12 and 28 result from calculation or operation errors, while 3 comes from subtracting the common difference. Exam tip: for the second term, use \(a+d\) directly.