Introduction to APs and common difference.

In option A, the differences between consecutive terms are \(53-60=-7\), \(46-53=-7\), and \(39-46=-7\). Since every difference is the same, it is an arithmetic progression with \(d=-7\). Option B has common difference \(+7\), while options C and D do not have equal consecutive differences. Exam tip: To identify an AP, check the difference between every pair of consecutive terms.

(\frac{7}{6}-\frac{5}{6}=\frac{1}{3}) and (\frac{3}{2}-\frac{7}{6}=\frac{1}{3}). For fractions subtract using common denominators.

In an arithmetic progression, each new term is obtained by adding the common difference d to the previous term. Here, a = -9 and d = 6: -9, -9+6 = -3, -3+6 = 3, and 3+6 = 9. Therefore, the first four terms are (-9, -3, 3, 9). Option D subtracts 6 each time, so its common difference is -6. Exam tip: check the difference between consecutive terms to verify an AP.

The common difference is found by subtracting a term from the next term. Here, \(1.9-1.4=0.5\) and \(2.4-1.9=0.5\). Since the difference between consecutive terms is the same, the common difference is \(0.5\). Option \(0.4\) is a distractor because it is not the difference between consecutive terms. Exam tip: Check at least two consecutive differences to confirm an AP.

In an arithmetic progression, the common difference \(d\) equals the later term minus the earlier term. Therefore, \(d=35-27=8\). \(-8\) would result only if the terms were in the order \(35, 27\). Exam tip: For two consecutive terms, subtract the first term from the second term.

For consecutive terms of an arithmetic progression, the difference between adjacent terms is constant. From the first two terms, the common difference is \(d=11-5=6\). Add this difference to the second term to obtain the third term: \(n=11+6=17\). Adding 6 once more gives \(17+6=23\), matching the fourth term and confirming the calculation.

Therefore \(n=17\), so option C is correct. The sequence is \(5,11,17,23\), and each term increases by 6. Values such as 15 or 16 would make the differences unequal, while 18 would produce a third-to-fourth difference of only 5. Hence only 17 satisfies the arithmetic-progression condition.

The direct answer is option B: \((14,18,22,26,\ldots)\). In an arithmetic progression, \(a\) means the first term and \(d\) means the common difference. We need both conditions: first term must be 14, and each next term must increase by 4. Check option A: its first term is 4, although its difference is 4, so it fails the first condition. Check option B: the first term is 14, and \(18-14=4\), \(22-18=4\), and \(26-22=4\). Both conditions are satisfied, so B is correct. Check option C: its first term is 14, but its common difference is \(17-14=3\), not 4. Check option D: its difference is 4, but its first term is 18 rather than 14. Therefore only option B has \(a=14\) and \(d=4\). A useful formula is \(a_n=a+(n-1)d\), but no formula is needed here because the first two terms show the required values directly. The common mistake is checking only the difference or only the first term. Memory cue: for an AP option, inspect the first term for \(a\), then subtract consecutive terms for \(d\).

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 direct answer is option B: \(d=1.5\) GB. In an arithmetic progression, the common difference \(d\) is the fixed amount added to one term to obtain the next term. Here the data values are 2, 3.5, 5, and 6.5. Calculate successive differences: \(3.5-2=1.5\), \(5-3.5=1.5\), and \(6.5-5=1.5\). Since the difference is the same each time, the sequence is an AP and its common difference is 1.5 GB. Option A, 1 GB, is too small and does not turn 2 into 3.5. Option B, 1.5 GB, exactly matches every increase. Option C, 2 GB, would make the next term 4 rather than 3.5. Option D, 3.5 GB, is the second term, not the increase between terms. The word “each day” tells us to compare consecutive days, not to choose a term itself. Memory cue: common difference equals “next term minus previous term”; check it in at least two places.

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.

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.