Arithmetic Progression

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.

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.

The direct answer is option C, 27. This is an arithmetic sequence because the same number is added each time. Subtract consecutive terms: \(12-7=5\), \(17-12=5\), and \(22-17=5\). Therefore the common difference is 5. The next term is \(22+5=27\). Option C is correct because it follows the established pattern. Option A, 25, would add only 3 to 22, so it breaks the common difference. Option B, 26, would add 4, also incorrect. Option D, 28, would add 6, so it does not continue the sequence. A weak student should not guess from the last number; first compare each neighbouring pair. In an arithmetic progression, the gap between consecutive terms remains constant. The formula \(a_n=a_1+(n-1)d\) also confirms the pattern, with \(a_1=7\) and \(d=5\). Memory cue: find the repeated difference, then add it once to the final displayed term.

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, subtracting any term from the following term gives the same constant difference. Here, 18 − 20 = −2, 16 − 18 = −2, and 14 − 16 = −2, so the common difference is d = −2. Continue the rule from the last displayed term: 14 + (−2) = 12. Therefore option C is correct. Option A subtracts 4 instead of 2, option B does not follow the established constant difference, and option D increases the sequence rather than continuing its decrease. A negative common difference is valid and simply indicates a decreasing arithmetic progression.

The governing concept is the constant common difference of an arithmetic progression. Subtract consecutive terms: 10−6=4, 14−10=4, and 18−14=4. Hence the common difference is d=4. The displayed sequence contains the first four terms, so the fifth term is obtained by adding the same difference to the fourth term: a₅=18+4=22. Therefore option C is correct. The result can also be checked using aₙ=a₁+(n−1)d: a₅=6+(5−1)4=6+16=22. Options A and B do not continue the established increase of four, while option D would require an increase of six from 18. Identifying the constant difference avoids guessing and confirms the unique answer.

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.

The governing definition is that an arithmetic progression has a constant difference between consecutive terms. For this sequence, the differences are 4 − 2 = 2, 8 − 4 = 4 and 16 − 8 = 8. Since 2, 4 and 8 are not equal, the sequence fails the defining condition and is not an AP. In fact, its terms are obtained by multiplying by 2, which is characteristic of a geometric pattern rather than an arithmetic one. Therefore option C is correct. Increasing terms can still form an AP, the first term may be any suitable number, and being even has no role in deciding whether the differences are equal.

In an arithmetic progression, the common difference d equals the second term minus the first term. Here, d = 7 − 3 = 4. Also, 11 − 7 = 4, confirming that the difference is constant. The number 3 is the first term, not the common difference. Exam tip: subtract any term from the next consecutive term to find d.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_4=12+(4-1)\times5=12+15=27\). The value 24 is the third term because it is obtained by adding the common difference only twice. Exam tip: for the fourth term, use \(a+3d\).

The difference is (-5), so the terms are (30,25,20,15,10,5). In exams, include the first term while counting.

Direct answer: option C, the fifth term. An arithmetic progression has a constant difference between consecutive terms. Here 6−1 = 5, 11−6 = 5, and 16−11 = 5, so the common difference is d = 5. Continuing once more gives 16 + 5 = 21, which immediately shows that 21 is the fifth term. Using the formula confirms it: aₙ = a₁ + (n−1)d. Set 21 = 1 + (n−1)5. Subtract 1: 20 = 5(n−1). Divide by 5: 4 = n−1, so n = 5. A, the third term, is 11; B, the fourth term, is 16; and D, the sixth term, would be 26. Thus only C is correct. Memory cue: in an AP, count terms from the first listed term and add the same common difference each time.

Direct answer: option A, aₙ = 5n + 3. The first term is a₁ = 8, and the common difference is d = 13−8 = 5. For an arithmetic progression, the nth-term formula is aₙ = a₁ + (n−1)d. Substitute the known values: aₙ = 8 + (n−1)5 = 8 + 5n − 5 = 5n + 3. Thus A is correct. Check it directly: n = 1 gives 8, n = 2 gives 13, and n = 3 gives 18. Option B gives the first term 8 but its difference is 8, not 5. Option C gives 2 when n = 1, so it does not start correctly. Option D gives 8 initially but increases by 1, not by 5. A valid AP rule must have both the correct first term and the correct constant difference. Common warning: matching only the first term is not enough.

The common difference is (6), so the next term is (16+6=22). In exams, continue the same difference for the blank term.

The common difference of an arithmetic progression is the fixed amount obtained when one term is subtracted from the next term. It may be positive when the terms increase, negative when they decrease, or zero when they remain constant. Here the sequence moves downward from 11 to 8, then to 5 and 2, so the difference should be negative.

Using consecutive terms, subtract the first term from the second: \(8-11=-3\). Checking the next pairs gives \(5-8=-3\) and \(2-5=-3\), confirming that the difference is constant. Therefore the correct choice is B, \(-3\). Choice A, 3, gives the size of the decrease but not its signed common difference.

The governing concept is the definition of an arithmetic progression: the difference between every pair of consecutive terms must be constant. For this sequence, 20 − 10 = 10, 30 − 20 = 10, and 40 − 30 = 10. Since the same difference occurs each time, the sequence is an arithmetic progression with common difference d = 10. Thus, option A is correct. Option B incorrectly uses 20, which is a term rather than the difference between consecutive terms. Option C is wrong because an arithmetic progression does not require all terms to be equal; it requires their differences to be equal. Option D is wrong because the sequence increases, rather than decreases. The repeated difference provides the decisive test.