Arithmetic Progression

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 governing concept is the common difference in an arithmetic progression. Calculate each consecutive difference: 31 − 24 = 7, 38 − 31 = 7, and 45 − 38 = 7. Since the difference remains constant, the next term must be obtained by adding 7 to the last known term. Thus the next term is 45 + 7 = 52, making option C correct. A recursive description of the same rule is: start with 24 and add 7 repeatedly, producing 31, 38, 45, and then 52. Option A would add 5, option B would add 6, and option D would add 8, so each would violate the established pattern. The repeated equal difference proves that the answer is unique rather than merely a visual guess.

The direct answer is option B, 100. In an AP, the common difference remains constant. Here each term is 10 less than the preceding term: 130−140=−10, 120−130=−10, and 110−120=−10. The listed terms are first 140, second 130, third 120, and fourth 110. Therefore the fifth term is 110−10=100. Option A, 90, is the sixth term, not the fifth. Option B, 100, is obtained after one further subtraction and is correct. Option C, 110, is already the fourth term, so it stops one step too early. Option D, 120, is the third term and is even earlier. Always count positions carefully rather than merely continuing until a familiar number appears. Memory cue: for the fifth term, make four moves from the first term: 140−4(10)=100.

In an arithmetic progression, the common difference is the fixed amount added to one term to obtain the next term. It can be found by subtracting any term from the term immediately after it. The letter p represents an algebraic value, but it cancels during subtraction, so the answer does not depend on the value of p.

Using the first two terms, the difference is \\( (p+7)-p=7 \\). The same result appears again because \\( (p+14)-(p+7)=7 \\). Thus every consecutive pair differs by 7, so the common difference is 7. Therefore, option B is correct. Option C is the increase from the first term to the third term, not the common difference between consecutive terms.

Consecutive terms are equal, so the difference is (11-11=0). In exams, a constant sequence can also be an arithmetic progression.

In an arithmetic progression, the first term is represented by \(a\), and the common difference \(d\) is added to obtain each next term. If \(a=15\) and \(d=0\), no change occurs when moving from one term to the next. Thus the sequence remains constant rather than increasing or decreasing.

The first term is 15. The second term is \(a+d=15+0=15\), and the third term is \(15+0=15\) again. Therefore, the first three terms are \(15,15,15\), so option C is correct. The general term also confirms this: \(a_n=a+(n-1)d=15+(n-1)0=15\) for every positive integer \(n\). Choices A and B incorrectly change the position of 15, while choice D treats the difference as 15 instead of zero.

The difference between (18) and (25) is (7), so the missing term is (18-7=11). In exams, apply the common difference backward too.

In the linear rule (a_n=9n-2), the coefficient of (n) is the difference (9). In exams, take (k) as (d) in (kn+c).

The first term is 18 and the common difference is 5. Therefore, the sixth term is \(a_6=a+(6-1)d=18+5\times5=43\). Choosing 40 results from adding the common difference only four times. Exam tip: use \(a_n=a+(n-1)d\) for the nth term of an AP.

The nth term of an arithmetic progression is \(a_n=a_1+(n-1)d\). Therefore, \(a_4=13+(4-1)\times 9=13+27=40\). Hence, \(40\) is correct. \(43\) would result from adding the common difference four times, but only three differences are added to reach the fourth term. Exam tip: use \(n-1\), not \(n\), when finding \(a_n\).

In an arithmetic progression, the difference between consecutive terms, called the common difference, is constant; for example, 3, 8, 13 has difference 5. A geometric progression has a constant ratio, not a constant difference. Exam tip: “constant difference” identifies an AP.

The direct answer is option B, 110. The sequence is 11, 22, 33, 44, and so on. Each term increases by 11: 22−11=11, 33−22=11, and 44−33=11. It is also the sequence of positive multiples of 11. The nth term is therefore 11n, so the tenth term is 11×10=110. Option A, 99, is 11×9 and is the ninth term, so it is one position early. Option B, 110, is 11×10 and is correct. Option C, 121, is 11×11, the eleventh term, so it is one position late. Option D, 132, is 11×12, the twelfth term, and is also too large. The important point is to multiply 11 by the term number, not by a number chosen from the displayed list. Memory cue: in 11, 22, 33, the nth term is 11n.

Given \(a_n=18-4n\), substitute \(n=3\): \(a_3=18-4(3)=18-12=6\). Hence, 6 is the correct option. A value such as 8 can result from an incorrect calculation of \(4\times3\). Exam tip: while finding a term, substitute the value of \(n\) carefully into the complete expression.

At (n=1) it gives (14), and at (n=2) it gives (10), so (a_n=18-4n). In exams, choose the rule after checking the first term.

The middle term is the average of (12) and (30), so (x=\frac{12+30}{2}=21). In exams, take the average for the middle term in three-term progressions.

Here, the first term is \(a=30\) and the common difference is \(d=38-30=8\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_7=30+(7-1)\times 8=30+48=78\). Option 76 would result from adding the difference only five times, but the seventh term requires six additions of the common difference. Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

An arithmetic progression is identified by two details: its first term and the constant change from one term to the next. The required sequence must begin with 12, and every new term must be obtained by adding 6. Checking both conditions prevents confusion with a sequence that has the right difference but the wrong first term, or the right first term but a different difference.

In option A, the sequence begins 12, and its consecutive differences are \\(18-12=6\\), \\(24-18=6\\), and \\(30-24=6\\). Thus its first term is 12 and its common difference is 6. Option B begins with 6, option C increases by 12, and option D begins with 18. Therefore, option A is the only correct choice.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_6=10+(6-1)\times7=10+35=45\). The value 42 is obtained by adding the common difference only four times, so it is the fifth term. Exam tip: always subtract 1 from the term number before multiplying by \(d\).

An arithmetic progression must have a constant difference between consecutive terms, and the fourth listed term must be 20. In option A, the differences are 8 − 2 = 6, 14 − 8 = 6, and 20 − 14 = 6. Thus it is an AP with common difference 6, and its fourth term is exactly 20. Option B also has common difference 6, but its fourth term is 22, so it fails the first condition. Option C has fourth term 26, while option D has fourth term 38. Therefore, only option A satisfies both conditions simultaneously. The governing idea is that checking the common difference alone is insufficient; the specified term position must also be checked.

The governing property of an arithmetic progression is a constant difference between consecutive terms. For three numbers p, q, r to be consecutive AP terms, q - p must equal r - q; equivalently, 2q = p + r. In option B, 12 - 6 = 6 and 18 - 12 = 6, so the differences are equal and the numbers can be consecutive terms of an AP. Option A has differences 4 and 5, option C has differences 5 and 8, and option D has differences 6 and 9. Since the two successive differences are unequal in each of those cases, those options cannot represent three consecutive arithmetic-progression terms. Thus option B is the only valid choice.

Use the standard arithmetic-progression relation a_n = a_1 + (n - 1)d. For the fifth term, substitute a_5 = 30, n = 5, and d = 7: 30 = a_1 + (5 - 1)7 = a_1 + 28. Subtracting 28 from both sides gives a_1 = 2, so option A is correct. The result can also be obtained by moving backwards four equal steps from the fifth term: 30, 23, 16, 9, 2. Four steps are required because the index changes from 5 to 1 by 4, not by 5. Option B would subtract only three differences and reach the second term, while options C and D likewise fail to move back the correct number of steps.

The relevant concept is the nth-term formula for an arithmetic progression: aₙ = a₁ + (n − 1)d. In the given progression, the first term is x and the common difference is 8. From the first term to the fourth term there are three equal intervals, so the fourth term is x + 3(8), not x + 4(8). Since the fourth term equals 41, x + 24 = 41, and therefore x = 17. Thus option B is correct. Direct substitution confirms the result: the progression becomes 17, 25, 33, 41. If x were 15, 19, or 21, the fourth term would be 39, 43, or 45 respectively. The essential reasoning is to count the number of differences between term one and term four correctly.

The governing rule is aₙ = a₁ + (n − 1)d. For the fourth term, three common differences separate the first term from it, so a₄ = a₁ + 3d. Substituting the given information gives 17 = a₁ + 3(4) = a₁ + 12. Therefore a₁ = 17 − 12 = 5, so option A is correct. This can be verified by working backward from 17: subtracting the common difference 4 gives the third term 13, then the second term 9, and finally the first term 5. If the first term were 6, 7, or 9, adding 12 would give fourth terms 18, 19, or 21 rather than 17. The key is that the fourth term involves exactly three differences.

The governing concept is the nth-term formula aₙ = a₁ + (n − 1)d. For the sixth term, five common differences separate a₁ from a₆, so a₆ = a₁ + 5d. Substituting the given values gives 41 = a₁ + 5(8) = a₁ + 40. Subtracting 40 from both sides yields a₁ = 1, so option A is correct. A backward check confirms the result: starting with 41 and subtracting 8 successively gives the fifth term 33, fourth term 25, third term 17, second term 9, and first term 1. If a₁ were 3, 5, or 7, the sixth term would be 43, 45, or 47 respectively, not 41. Thus the first term is uniquely determined as 1.