Arithmetic Progression

Each term decreases by (5), so the common difference is (-5). In exams, write the difference as negative for a decreasing progression.

A sequence is an arithmetic progression when the same number is added to obtain each next term. That fixed number is called the common difference. It is enough to compare consecutive terms, but checking all displayed pairs makes the pattern clear and prevents confusing the first term with the difference.

Here, \(13-8=5\), \(18-13=5\), and \(23-18=5\). Every difference is equal, so the sequence is an arithmetic progression with common difference 5. Hence choice A is correct. Choice B gives the first term rather than the change between terms. Choices C and D are incorrect because the differences are equal and positive, showing that the terms increase regularly.

In an arithmetic progression, the difference between every pair of consecutive terms must be the same. Here, the differences are \(4-2=2\), \(7-4=3\), and \(11-7=4\). Since these differences are unequal, the sequence is not an arithmetic progression. Increasing or positive terms alone do not make a sequence an AP. Exam tip: Find two or three consecutive differences to check whether a sequence is an AP.

An arithmetic progression is a sequence in which the difference between consecutive terms remains constant. However, to identify its first term, we do not need to calculate that difference. The terms are written in their order, and the first number displayed is the first term. In the sequence 16, 22, 28, 34, the number 16 appears first, so the first term is 16. Therefore, choice B is correct.

The common difference can be checked as \(22-16=6\), \(28-22=6\), and \(34-28=6\), confirming that the sequence is an arithmetic progression. This difference does not change the identity of the first term. The numbers 22 and 28 are later terms, while 6 is only the common difference, not a term in the displayed progression.

The general term of an arithmetic progression is


a_n=a+(n-1)d
a

. For the third term,


a_3=7+(3-1) imes5=7+10=17
a

. Therefore, 17 is correct. The value 15 is only the second term because the common difference 5 is added once. Exam tip: for the third term, add the common difference twice to the first term.

The direct answer is option C, 26. An arithmetic progression (AP) is a sequence in which the same number is added or subtracted each time. Compare consecutive terms: 38−42=−4, 34−38=−4, and 30−34=−4. Thus the common difference is −4. The next term is 30+(−4)=26. Option A, 22, would result from subtracting 8, which is not the observed difference. Option B, 24, would result from subtracting 6, so it does not follow the pattern. Option C, 26, correctly continues the repeated subtraction of 4. Option D, 28, would result from subtracting only 2. Since the terms decrease by 4 at every step, 26 is certain. Memory cue: in a decreasing AP, write the common difference with its negative sign and apply it once more.

The difference between (21) and (28) is (7), so the missing term is (21-7=14). In exams, apply the difference backward too.

This is a sequence of multiples of (8), so (a_n=8n). In exams, identify multiple-based progressions as (kn).

Given \(a_n=5n+1\), substitute \(n=6\) for the sixth term: \(a_6=5\times 6+1=30+1=31\). Therefore, 31 is correct. The value 29 could result from subtracting 1 instead of adding it, so it is not correct. Exam tip: For a term \(a_n\), first substitute the value of \(n\), then perform multiplication and addition in order.

Here, the first term is \(a=9\) and the common difference is \(d=13-9=4\). Thus, \(a_n=a+(n-1)d=9+4(n-1)=4n+5\). Therefore, option B is correct. In option C, putting \(n=1\) gives the first term as \(-1\), not 9. Exam tip: identify the first term and common difference, then apply \(a_n=a+(n-1)d\) directly.

In an arithmetic progression, the second term equals the first term plus the common difference. Thus, the second term is 22 + (-6) = 16. Getting 18 would mean subtracting 4 from 22, but the given difference is 6. Exam tip: a negative common difference means that successive terms decrease.

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.

An arithmetic progression changes by the same amount from one term to the next. Here, each term increases by 7: from 3 to 10, from 10 to 17, and from 17 to 24. Therefore, the terms can be continued by adding 7 each time. The question asks for the position of 38, not merely whether 38 belongs to the progression.

Starting with the first term, the sequence is 3, 10, 17, 24, 31, 38. Counting carefully, 3 is the first term, 10 the second, 17 the third, 24 the fourth, 31 the fifth, and 38 the sixth. The formula gives the same result: using the first term and common difference, solve \(3+(n-1)7=38\), so \(n-1=5\) and \(n=6\). Hence, option C is correct. Option B stops at 31, while option D would be 45.

The answer is option A, \(a_n=5n+10\). The first term is 15 and the common difference is \(20-15=5\). The general AP formula is \(a_n=a+(n-1)d\). Substituting gives \(a_n=15+(n-1)5=15+5n-5=5n+10\). Checking \(n=1\) gives 15, \(n=2\) gives 20, and \(n=3\) gives 25, so the formula fits. Option A is correct. Option B, \(15n\), gives 15, 30, 45, so its difference is 15. Option C, \(5n-10\), gives -5 for the first term and is wrong. Option D, \(n+14\), gives 15 first but increases by only 1, not 5. Do not confuse the first term with the coefficient of n. Memory cue: use \(a_n=a+(n-1)d\).

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.

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.