Arithmetic Progression

In an arithmetic progression, the difference between every pair of consecutive terms must be the same. Here, the differences are 10−5=5, 20−10=10, and 40−20=20. Since these differences are unequal, this is not an arithmetic progression. Merely increasing terms do not make a sequence an AP. Exam tip: check the first two or three consecutive differences.

In an arithmetic progression, the common difference d is the difference between consecutive terms. Here, d = 15 - 6 = 9. Checking further, 24 - 15 = 9 as well, so the correct answer is 9. The number 15 is the second term, not the common difference. Exam tip: find d by subtracting the previous term from the next term.

Here, 21−18=3, 24−21=3, and 27−24=3. Equal differences between consecutive terms confirm an AP, so its common difference is 3. Exam tip: check at least two differences.

The consecutive difference is constant: 60 − 65 = −5, 55 − 60 = −5, and 50 − 55 = −5. Hence d = −5. Starting with the first term, the sequence continues as 65, 60, 55, 50, 45, 40, so the sixth term is 40. Using the formula gives a₆ = a₁ + (6 − 1)d = 65 + 5(−5) = 40. Thus option B is correct.

The terms are (3,10,17,24,31,38), so (38) is the sixth term. In exams, list terms in order to check.

Putting (n=1,2,3) gives (10,14,18). In exams, use small (n) values for the first three terms.

The first term is (15) and the difference is (5), so (a_n=5n+10). In exams, find the constant part from the first term.

The common difference is (9), so the next term is (26+9=35). In exams, continue the same difference for the blank term.

The governing concept is the signed common difference of an arithmetic progression. It is found by subtracting a term from the term immediately after it: d = next term − previous term. Using the first pair, d = 21 − 27 = −6. The next pairs confirm the same value: 15 − 21 = −6 and 9 − 15 = −6. Thus each step decreases the sequence by 6, so the common difference, including its sign, is −6. Option B is correct. Option A gives only the positive magnitude and ignores that the sequence is decreasing. Option C is a listed term rather than a difference, while option D has the wrong magnitude. Keeping the sign is essential when describing an arithmetic progression.

(a=18) and (d=7), so (a_8=18+7\times7=67). In exams, use (a+(n-1)d) for the (n)th term.

An arithmetic progression is a sequence in which the difference between every pair of consecutive terms is constant. For the given sequence, 24 − 12 = 12, 36 − 24 = 12, and 48 − 36 = 12. Since the same difference occurs at every step, it is an arithmetic progression with common difference 12. Therefore option A is correct. Option B incorrectly treats the second term as the difference. Option C confuses an arithmetic progression with a constant sequence; its terms do not need to be equal, only their consecutive differences must be equal. Option D is false because the sequence increases by 12 rather than decreases. The repeated subtraction check is sufficient to establish the classification.

In an arithmetic progression, the first term is a, and the common difference d is added to obtain each next term. Here, a = 5 and d = 11: 5, 5 + 11 = 16, 16 + 11 = 27, 27 + 11 = 38. Therefore, the correct sequence is 5, 16, 27, 38. In option C, 11 is incorrectly taken as the first term, whereas the first term is 5. Exam tip: check that the difference between every pair of consecutive terms is d.

At (n=1) it gives (48), and at (n=2) it gives (40), so (a_n=56-8n). In exams, check a decreasing rule with the first term.

Given \(a_n=56-8n\). Substituting \(n=4\), we get \(a_4=56-8(4)=56-32=24\). Hence, \(24\) is correct. \(26\) would result from an incorrect calculation instead of using \(8\times4=32\). Exam tip: Substitute the value of \(n\) first, then perform multiplication and subtraction in order.

In an arithmetic progression, \(a\) is the first term, so \(a=6\). The common difference \(d\) is the difference between consecutive terms: \(d=17-6=11\). This is confirmed by \(28-17=11\) and \(39-28=11\). In option D, \(17\) is the second term, not the first term. Exam tip: identify the first term as \(a\), then subtract consecutive terms to find \(d\).

In an arithmetic progression, subtracting each term from the next gives the same common difference; this is its defining property. A constant ratio indicates a geometric progression. Exam tip: check consecutive differences first.

This is an arithmetic progression with first term 8 and common difference 5. Its terms are 8, 13, 18, 23, 28, 33; therefore, 33 is the sixth term. The fifth term is 28, so it is not correct. In exams, start counting the terms from the first term as 1.

In an arithmetic progression, the common difference is the difference between consecutive terms. Thus, \(d=a_2-a_1=23-14=9\). Therefore, 9 is correct. Choosing 8 does not give the actual difference between the two terms. Exam tip: when the first two terms are given, calculate second term minus first term.

The common difference is (7), so the next term is (45+7=52). In exams, add the difference to the last known term.

The common difference is (-10), so the fifth term is (100). In exams, count in the correct direction for decreasing terms.