Introduction to APs and common difference.

In an arithmetic progression, the common difference is the same between every pair of consecutive terms. The first two terms here are 6 and 10, so \(d=10-6=4\). The third term is obtained by adding this difference to the second term: \(n=10+4=14\). The fourth term then becomes \(14+4=18\), which confirms that the sequence is consistent.

Therefore \(n=14\), so option D is correct. Another way to see this is that the middle term of four consecutive terms is equally spaced: the terms move from 6 to 10, then to 14, and then to 18. The values 12, 13, and 15 do not preserve the same difference throughout the progression.

For an arithmetic progression, a denotes the first term and d is the difference between any two consecutive terms. We must check both conditions, not just the starting number. In option B, the first term is 9 and the successive differences are 12 - 9 = 3, 15 - 12 = 3, and 18 - 15 = 3. Therefore it satisfies a = 9 and d = 3. Option A has d = 3 but starts with 3; option C starts with 9 but has d = 2; option D has d = 3 but starts with 12. Hence option B is the only correct choice.

The third term is (36) and the fourth is (33), so (33-36=-3). While finding the difference subtract the previous term from the next term.

In an arithmetic progression, the second term is obtained by adding the common difference to the first term: \(a+d=2.5+1.5=4.0\). Therefore, 4.0 is correct. The value 3.5 would result from adding only 1 to 2.5, not the given common difference of 1.5. Exam tip: For the second term of an AP, use \(a+d\) directly.

In the sequence (11,11,11,11,\ldots), every term is equal to the preceding term. Therefore, the common difference is \(d=11-11=0\). Option B has common difference \(-5\), not zero. Exam tip: To find the common difference of an AP, subtract the first term from the second term.

The first term is x and the second term is x+7. Thus, second term − first term = (x+7)-x=7. Therefore, the common difference is 7. Note that x+7 is the second term, not the difference. Exam tip: To find the common difference of an AP, subtract a term from the term immediately after it.

In this AP, the first term is \(a=16\) and the common difference is \(d=20-16=4\). Therefore, \(a+d=16+4=20\), which is the second term of the AP. \(24\) is the third term, so it is not correct. Exam tip: \(a+d\) is always the second term of an AP.

In an arithmetic progression, the difference between consecutive terms is constant. Here, the common difference is \(d=18-21=-3\). Therefore, the third term is \(q=18+(-3)=15\). Checking: \(15-3=12\), so \(21,18,15,12\) is correct. If \(14\) were used, the consecutive differences would not be equal. Exam tip: Find \(d\) from the first two terms and apply the same difference to obtain the next term.

Every consecutive difference is (5), so it is an arithmetic progression. While choosing the statement check both type and (d).

In an arithmetic progression, the common difference is d = a₂ − a₁. Thus, d = 8 − 13 = -5. Therefore, -5 is correct. Choosing 5 would reverse the order of subtraction. In exams, subtract the first term from the second term to find d.

In the general form (a,a+d,a+2d,\ldots), (a+2d) is the third term. Remembering the general form makes term identification easier.

This is an AP since 11−15=−4, 7−11=−4, and 3−7=−4. A negative common difference makes terms decrease but does not stop it from being an AP. Exam tip: compare consecutive differences.

In an AP, \(a\) is the first term and \(d\) is the difference between consecutive terms. In option B, the first term is \(-4\), and \(2-(-4)=6\) and \(8-2=6\). Hence, it has \(a=-4\) and \(d=6\). Option C also has common difference 6, but its first term is 4. Exam tip: check the first term first, then subtract the first term from the second term to find \(d\).

The common difference is (\frac{1}{2}), so the next term is (\frac{5}{2}+\frac{1}{2}=3). Keep denominators common when adding fractions.

The common difference of an AP is the difference between consecutive terms. Here, \((5+s)-5=s\) and \((5+2s)-(5+s)=s\), so the common difference is \(s\). \(2s\) is the difference between the first and third terms, not between consecutive terms. Exam tip: Find the common difference by subtracting the first term from the second term.

The equal hourly increase is (20) rupees, so (d=20). In word problems treat the equal increase as the common difference.

In the sequence \(20,17,14,11,\ldots\), the consecutive differences are \(17-20=-3\) and \(14-17=-3\). Therefore, its common difference is \(d=-3\), not \(-2\). In each of the other three sequences, every next term is 2 less, so \(d=-2\). Exam tip: find the common difference using \(d=a_2-a_1\).

The nth term of an AP is \(a_n=a+(n-1)d\). Therefore, \(a_4=10+(4-1)\times 0=10\). When \(d=0\), every term is equal to the first term; a value such as 20 would occur only if the common difference were non-zero. Exam tip: Substitute the term number \(n\) and common difference \(d\) carefully in the formula.

In this AP, the first term is \(a=3\) and the common difference is \(d=8-3=5\). Therefore, \(a+2d=3+2(5)=13\). Hence, 13 is the correct option. The number 8 is the second term, whereas \(a+2d\) represents the third term of an AP. Exam tip: use \(a+(n-1)d\) to find the \(n\)th term.

In an arithmetic progression, the common difference \(d\) is the difference between consecutive terms. Here, \((y+1)-(y-2)=y+1-y+2=3\). Also, \((y+4)-(y+1)=3\), so \(d=3\). \(y-2\) and \(y+1\) are terms of the AP, not the common difference. Exam tip: Find \(d=a_2-a_1\) and verify it using the next pair of terms.