Introduction to APs and common difference.

Find the common difference from consecutive terms: d=45−50=−5, and the same change appears in 40−45 and 35−40. To obtain the next term, subtract 5 from the last displayed term: 35+(−5)=30. The question asks for the next term and d in that order, so the pair is (30,−5). Hence option D is correct; option A has the wrong sign for d.

In an arithmetic progression, the first term is a, and each next term is obtained by adding the common difference d. Here, a=12 and d=-4, so the terms are 12, 12+(-4)=8, and 8+(-4)=4. Therefore, the correct order is (12,8,4). Option B uses a common difference of +4, while option C does not begin with the first term 12. Exam tip: A negative d means that the same value is subtracted from each successive term.

In an AP, \(a\) is the first term, so \(a=0\). The common difference \(d\) is the difference between consecutive terms: \(d=-3-0=-3\). Hence, \(a=0,\ d=-3\) is correct. In option A, the sign of the common difference is incorrect. Exam tip: use \(d=a_2-a_1\), subtracting the first term from the second term.

The AP is 20, 35, 50, 65, \ldots, where \(a_1=20\) and \(a_3=50\). Therefore, \(a_3-a_1=50-20=30\). The common difference is \(d=15\), but the difference between the first and third terms is \(2d=30\), so 15 is not correct. Exam tip: When asked for the difference between two terms, subtract the stated terms directly to verify your answer.

Adding the same (2) to each term does not change the difference, so (d=5). Equal change in all terms does not change (d).

The common difference is (-2-(-7)=5), so the third term is (-2+5=3). Watch signs carefully with negative numbers.

The common difference is found by subtracting a term from the next term: \((2a-5)-2a=-5\). Also, \((2a-10)-(2a-5)=-5\), confirming that the difference is constant. Each next term decreases by 5, so the common difference is \(-5\); \(5\) is only the magnitude of the decrease. Exam tip: always keep the sign while writing the common difference.

For three consecutive terms of an arithmetic progression, the middle term equals the average of the first and third terms. Thus, \(b=\frac{9+21}{2}=15\). Therefore, 15 is the correct option. For instance, if \(b=13\), the differences are \(13-9=4\) and \(21-13=8\), which are not equal. Exam tip: For three consecutive AP terms \(a-d,\ a,\ a+d\), the middle term is always the average of the outer terms.

In (\frac{1}{3},1,\frac{5}{3},\frac{7}{3},\ldots), every difference is (\frac{2}{3}). Check fractions using common denominators.

In option A, consecutive differences are equal: 9 − 4 = 5, 14 − 9 = 5, and 19 − 14 = 5. Hence it is an AP. The differences change in the other options. Exam tip: check at least two consecutive differences.

In an arithmetic progression, the common difference \(d\) is the difference between consecutive terms. Here, \(d=6.8-6=0.8\). Checking further, \(7.6-6.8=0.8\) and \(8.4-7.6=0.8\). The value 0.6 would result from incorrect decimal subtraction. Exam tip: find \(d=a_2-a_1\) and verify it using the next pair of terms.

The common difference is next term minus previous term, so (10-15=-5). In a decreasing sequence the order of subtraction is very important.

In an arithmetic progression, the second term is obtained by adding the common difference to the first term: \(a_2=a_1+d\). Thus, \(11=a_1+4\), so \(a_1=7\). Option 11 is the second term, not the first term. Exam tip: subtract \(d\) from the second term to find the first term.

The difference from (18) to (24) is (6), so (u=18-6=12). While finding a missing term use the same difference forward and backward.

The common difference is the difference between consecutive terms. Here, \(13-8=5\) and \(18-13=5\), so 5 is added to each term to obtain the next term. Therefore, the common difference is 5. Although 4 is a close distractor, it is not the difference between any two consecutive terms here. Exam tip: For an AP, quickly verify using \(d=a_2-a_1\).

The common difference is found by subtracting a term from the next term: \(36-42=-6\). Also, \(30-36=-6\) and \(24-30=-6\), so the common difference is \(-6\). Option 6 ignores the sign; in a decreasing AP, the common difference is negative. Exam tip: always calculate \(a_2-a_1\) in this order to get the correct sign.

This is an arithmetic progression because the differences 10-6 and 14-10 are both 4. Therefore, the term after 14 is 14+4=18. Checking further, 18+4=22. Option 16 would give a difference of 2 from 14, so it is not correct. Exam tip: In an AP missing-term question, verify that the common difference is the same on both sides of the blank.

In an arithmetic progression, each term is obtained by adding the common difference to the preceding term. Starting with 12 and adding 5 repeatedly gives 12, 12+5=17, 17+5=22, and 22+5=27. Therefore, (12,17,22,27) is correct. In option C, the common difference is 6, not 5. Exam tip: check the difference between consecutive terms before selecting an option.

In an AP, \(a\) is the first term, so \(a=-5\). The common difference \(d\) is the difference between consecutive terms: \(d=-1-(-5)=4\). Hence, the correct pair is \(a=-5,\ d=4\). In option C, the common difference is correct, but the first term is not. Exam tip: To find \(d\), subtract the previous term from the next term.

The difference between consecutive terms is constant: \(27-31=-4\), \(23-27=-4\), and \(19-23=-4\). Hence, the common difference is \(d=-4\). Therefore, the next term is \(19+(-4)=15\). Choosing 16 would give a difference of \(-3\), which does not match the common difference of the progression. Exam tip: To find the next term, add the common difference to the last term.