Arithmetic Progression

The governing pattern is an arithmetic progression with first term 1 and common difference 3. The rule x=3n+1, with n∈W and 0≤n≤4, produces x=1,4,7,10,13 when n=0,1,2,3,4. Thus it gives exactly the five members of B. Option B produces multiples of 3, option C has difference 2, and option D has difference 4, so option A is correct.

A sequence has a common difference when the same number is added to obtain each next term. Subtract consecutive terms: 16 - 11 = 5, 21 - 16 = 5, and 26 - 21 = 5. Since the difference remains constant, the sequence is arithmetic and its common difference is 5. Therefore option A is correct; the other numbers do not reproduce the given terms.

The governing concept is the arithmetic progression, whose first term is 5 and common difference is 7. Its nth term is a_n = 5 + (n - 1)7 = 7n - 2. We need 50 < 7n - 2 < 150. Adding 2 gives 52 < 7n < 152, and division by 7 gives 7.43... < n < 21.71.... Therefore the integer values of n are 8 through 21. The number of integers in this inclusive range is 21 - 8 + 1 = 14. The corresponding first and last terms are 54 and 145, both satisfying the strict inequalities. Thus option B is correct; options A, C, and D result from miscounting one or more endpoints or treating 50 or 150 as included.

The sequence is geometric: the first term is 5 and the common ratio is 2. Therefore its nth term is a_n = 5×2^(n-1), because the first term contains 2^0, the second contains 2^1, and so on. To test 5120, divide by 5: 5120 ÷ 5 = 1024 = 2^10. Thus 5120 = 5×2^10. Comparing exponents, n - 1 = 10, so n = 11. Hence option B is correct. An answer of 10 would incorrectly use n rather than n-1; 12 or 13 would add too many doublings.

For an arithmetic sequence, the explicit rule is a_n=a_1+(n-1)d. From the first to the fourth term there are three equal gaps, so 3d=a_4-a_1=20-8=12, giving d=4. Substitute this into the formula: a_n=8+(n-1)4=8+4n-4=4n+4. Hence option B is correct. Option A gives a_1=12 rather than 8. Option C gives a_1=8 but a_4=23, so its difference is wrong. Option D gives a_1=8 but a_4=17. The key is dividing the total change by three gaps, not by four terms.

In an arithmetic progression, the common difference is found by subtracting a term from the next term. Here, \(5-2=3\). Also, \(8-5=3\) and \(11-8=3\), so the common difference is 3. The number 2 is the first term, not the common difference. Exam tip: subtract an earlier term from the following consecutive term.

(5) is added each time, so the next term is (22+5=27). In exams, first identify the common difference.

In an arithmetic progression, the common difference is added to a term to obtain the next term. Thus, second term = first term + common difference = \(4+6=10\). Therefore, 10 is correct. 12 is not the second term; it is the third term, \(4+2\times6\). Exam tip: use \(a_2=a_1+d\) for the second term of an AP.

Each term decreases by (4), so the common difference is (-4). In exams, a decreasing sequence has a negative difference.

An arithmetic progression is a sequence in which the difference between every pair of consecutive terms is the same. The common difference is found by subtracting each term from the term immediately after it. The actual size of the terms or whether they increase is not the defining test; equal consecutive differences are.

For this sequence, 6−3=3, 9−6=3, and 12−9=3. Since every checked difference is 3, the sequence is an arithmetic progression with common difference 3. Therefore option A is correct. The number 6 is a term, not the common difference, and the sequence is increasing rather than decreasing. The continuation dots do not change this conclusion.

The consecutive differences in this sequence are \(4-1=3\), \(9-4=5\), and \(16-9=7\). Since these differences are not equal, the sequence is not an arithmetic progression. An arithmetic progression need not begin with zero, and its terms do not all have to be positive. Exam tip: subtract each term from the next one; the sequence is an AP only when every consecutive difference is the same.

The governing concept is the identification of the first term of an arithmetic progression. In a written progression, the first term is the number that appears at the far left, usually denoted by a. Here the sequence begins with 9, so a = 9 and option B is correct. The common difference is 14 − 9 = 5, but that value is not the first term; it is the amount added to move from one term to the next. Option C is the second term, option D is the third term, and option A is the common difference. Reading the position of each displayed number carefully prevents confusing the first term with the difference.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). For the third term, \(n=3\), so \(a_3=5+(3-1)\times2=9\). Therefore, 9 is correct. 7 is only the second term because the common difference has been added once. Exam tip: to find the third term, add the common difference twice to the first term.

The governing concept is the common difference of an arithmetic progression. In an AP, the difference between every pair of consecutive terms remains constant. Here, 18 − 20 = −2, 16 − 18 = −2 and 14 − 16 = −2, so the common difference is d = −2. To obtain the next term, subtract 2 from the last displayed term: 14 + (−2) = 12. Therefore option C is correct. Option A subtracts 4 instead of 2, option B does not preserve the established difference, and option D increases the term. A negative common difference is completely valid and indicates that the progression is decreasing.

The governing concept is the constant common difference in an arithmetic progression. Starting from 6, the sequence increases by 4 each time: 10 − 6 = 4, 14 − 10 = 4 and 18 − 14 = 4. The displayed terms are therefore the first four terms, and the fifth term is found by adding the same common difference to the fourth term: a₅ = 18 + 4 = 22. Hence option C is correct. Option A and option B do not continue the difference of 4, while option D would require adding 6 to 18. Checking the difference before extending the sequence prevents these counting and calculation errors.

The difference between (8) and (11) is (3), so the blank term is (2+3=5). In exams, identify the difference from nearby known terms.

This is a sequence of multiples of (4), so (a_n=4n). In exams, identify multiple-based progressions as (kn).

To find the fourth term, substitute 4 for n in the formula: \(a_4=3\times4+1=12+1=13\). Therefore, the correct answer is 13. Option 12 is only \(3\times4\); the +1 in the formula has not been added. Exam tip: In any \(a_n\) formula, carefully substitute the required term number for n.

Here, the first term is \(a=5\) and the common difference is \(d=9-5=4\). Thus, \(a_n=a+(n-1)d=5+4(n-1)=4n+1\). In \(a_n=4n-1\), putting \(n=1\) gives 3 as the first term, so it is not correct. Exam tip: always substitute \(n=1\) in a general term to verify the first term.

In an arithmetic progression, the second term equals the first term plus the common difference. Thus, \(10+(-3)=7\). Option 13 would result if the common difference were \(+3\), but here it is negative. Exam tip: When adding a negative number, rewrite it as subtraction and check the sign carefully.