Arithmetic Progression

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.

The governing concept is locating a term in an arithmetic progression. The common difference is 5 because 6 − 1 = 5, 11 − 6 = 5 and 16 − 11 = 5. Continue the same pattern once more: the fifth term is 16 + 5 = 21. Equivalently, use aₙ = a + (n − 1)d, so 21 = 1 + (n − 1)5; then 20 = 5(n − 1), n − 1 = 4 and n = 5. Thus option C is correct. The third and fourth terms are 11 and 16, while the sixth term would be 26, so the neighbouring options are ruled out.

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.

In an arithmetic progression, the first term is 3 and 7 is added to obtain each next term: 3, 3+7=10, 10+7=17, 17+7=24. Therefore, the correct sequence is (3, 10, 17, 24). In option C, the first term has incorrectly been changed to 7. Exam tip: write the first term first, then add the common difference successively.

At (n=1) it gives (25), and at (n=2) it gives (20), so (a_n=30-5n). In exams, check a decreasing rule with the first term.

Given \(a_n=30-5n\). Substituting \(n=4\), we get \(a_4=30-5(4)=30-20=10\). Therefore, 10 is the correct option. Getting 15 would be incorrect because \(5\times4=20\), not 15. Exam tip: To find a particular term, substitute the given value of \(n\) carefully in the formula.

In an arithmetic progression, the first term is denoted by \(a\), so \(a=2\). The common difference \(d\) is the difference between consecutive terms: \(d=9-2=7\). Therefore, \(a=2,\ d=7\) is correct. In \(a=7,\ d=2\), the first term and common difference have been interchanged. Exam tip: find \(d\) by subtracting an earlier term from the next consecutive term.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Thus, \(a_5=6+(5-1)\times4=6+16=22\). Therefore, 22 is correct. The terms are 6, 10, 14, 18, 22, so 18 is the fourth term, not the fifth. Exam tip: To find the fifth term, add the common difference four times to the first term.

The terms are (4,7,10,13,16,19), so (19) is the sixth term. In exams, include the first term while counting.

In an arithmetic progression, the common difference is the difference between consecutive terms. Thus, \(d=a_2-a_1=14-8=6\). Therefore, 6 is the correct option. Choosing 5 does not give the correct subtraction of the first term from the second term. Exam tip: When the first two terms are given, use \(d=a_2-a_1\).

The governing concept is the common difference of an arithmetic progression. Subtract consecutive terms: 21 − 17 = 4, 25 − 21 = 4, and 29 − 25 = 4. Since the difference remains constant, the next term is found by adding 4 to the last displayed term: 29 + 4 = 33. Thus option C is correct. This is not a geometric progression, because the ratios are not constant; the question requires addition of the common difference, not multiplication by a common ratio. Option A is obtained by adding only 2, option B by adding 3, and option D by adding 5. None of those changes agrees with the repeated difference of 4 observed in every pair of consecutive terms.

An arithmetic progression is a sequence in which the difference between consecutive terms remains constant. In option C, 10 − 5 = 5, 15 − 10 = 5, and 20 − 15 = 5, so the common difference is 5. The differences in A double, those in B are 2, 3, and 4, and those in D are 3, 5, and 7. Therefore, only option C is an arithmetic progression.

The common difference is (-10), so the fifth term is (60). In exams, count in the correct direction for decreasing terms.

The governing concept is that the common difference is obtained by subtracting one term from the next. The first two terms are x and x + 4, so d = (x + 4) − x. Combining like algebraic terms gives d = x + 4 − x = 4. The same result appears from the next pair: (x + 8) − (x + 4) = 4. Therefore option C is correct, and the value does not depend on x. Option A incorrectly treats the variable part as the difference. Option D is the amount from the first term to the third term, not the difference between consecutive terms. Option B does not result from either subtraction and is therefore unsupported.

Consecutive terms are equal, so the difference is (3-3=0). In exams, a constant sequence can also be an arithmetic progression.