Introduction to APs and common difference.

In (14,18,22,26,\ldots), the first term is (14) and the difference is (4). Check both (a) and (d) together.

The third term is (63) and the fourth is (57), so the difference is (57-63=-6). While finding difference subtract the previous term from the next term.

The second term of an AP is obtained by adding the common difference to the first term: \(a_2=a+d=3.5+2.5=6.0\). Therefore, 6.0 is correct. Choosing 5.5 results from an incorrect addition of the decimals. Exam tip: use \(a_n=a+(n-1)d\); for the second term, put \(n=2\).

The first term is \(t\) and the second term is \(t+9\). Therefore, their difference is \((t+9)-t=9\). Hence, \(9\) is the correct answer. \(t+9\) is the second term, not the difference. Exam tip: In an AP, find the common difference using \(d=a_2-a_1\).

In this AP, the first term is \(a=21\) and the common difference is \(d=26-21=5\). Therefore, \(a+d=21+5=26\), which is the second term of the sequence. \(31\) is the third term, \(a+2d\), so it is not correct. Exam tip: In an AP, the second term is always \(a+d\).

In an arithmetic progression, the difference between consecutive terms is constant. Here, the common difference is \(d=42-48=-6\). Therefore, the third term is \(q=42+(-6)=36\). Checking further, \(36-6=30\), which matches the given next term. If 34 were chosen, the differences would be \(-6\) and then \(-8\), so it would not form an AP. Exam tip: For a missing-term question, verify the common difference using terms on both sides.

Every consecutive difference is (6), so it is an arithmetic progression. While choosing a statement check both type and (d).

In an arithmetic progression, the common difference is \(d=a_2-a_1\). Here, \(d=16-22=-6\), so the correct answer is \(-6\). Choosing \(6\) misses the negative sign; the second term is 6 less than the first term, so the common difference must be negative. Exam tip: always 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. Remember the general form to identify terms.

In an arithmetic progression, the difference between consecutive terms is constant. Here, 10 - 4 = 6 and 16 - 10 = 6, so the common difference is 6. Therefore, r = 16 + 6 = 22. Choosing 23 would give a difference of 7, which does not maintain the common difference. Exam tip: Find the difference between consecutive terms first, then add it to obtain the next term.

In \((-6,-1,4,9,\ldots)\), the first term is \(a=-6\). The difference between consecutive terms is \(-1-(-6)=5\) and \(4-(-1)=5\), so \(d=5\). Option C has common difference 5, but its first term is 6, so it is not correct. Exam tip: In an AP, check both the first term and the difference between consecutive terms.

The common difference is (1), so the next term is (\frac{11}{3}+1=\frac{14}{3}). Use a common denominator when adding an integer to a fraction.

The common difference of an arithmetic progression is the difference between two consecutive terms. Here, \((8+s)-8=s\). Also, \((8+2s)-(8+s)=s\), so the common difference is \(s\). The expression \(2s\) is the total added part in the third term, not the difference between consecutive terms. Exam tip: find the common difference by subtracting the preceding term from the next term.

The equal daily increase is (1.5), so (d=1.5). In word problems treat the equal increase as the common difference.

In the sequence \(45, 40, 35, 30, \ldots\), each term is 5 less than the preceding term. Therefore, its common difference is \(d=40-45=-5\), not \(-4\). In contrast, each of the other three sequences has common difference \(-4\). Exam tip: Find the common difference by subtracting the first term from the second term: \(d=a_2-a_1\).

In an AP, the difference between every pair of consecutive terms must be the same. Here, 9−5=4 and 13−9=4, but 18−13=5, so it is not an AP. Exam tip: check every consecutive difference, not just the first two.

In this AP, the first term is \(a=9\) and the common difference is \(d=16-9=7\). Therefore, \(a+2d=9+2\times7=23\). It is also the third term, since the third term of an AP is \(a+2d\). Option 30 is the fourth term, \(a+3d\), so it is not correct. Exam tip: The \(n\)th term of an AP is \(a+(n-1)d\).

In an arithmetic progression, the common difference is the difference between consecutive terms. Here, \(d=(z+2)-(z-4)=z+2-z+4=6\). Also, \((z+8)-(z+2)=6\), confirming that the common difference is 6. Option 4 is merely a number appearing in a term, not the difference. Exam tip: Find \(d\) first by subtracting the first term from the second term.

For an arithmetic progression, the common difference d is the same subtraction between every pair of consecutive terms. Here, 64−70=−6, 58−64=−6, and 52−58=−6, so d=−6. To obtain the next term, add this difference to the last known term: 52+(−6)=46. Therefore the next term and d, in the requested order, are 46 and −6, so option A is correct. Option B has the wrong sign for the difference and would make the sequence increase. Options C and D repeat terms already present rather than giving the next term. The negative difference correctly reflects that the sequence decreases by 6 each time.

The consecutive differences are 15−12=3, 18−15=3, and 21−18=3. Since all differences are equal, it is an AP with common difference 3. Exam tip: compare consecutive terms first.