Arithmetic Progression

This is an arithmetic progression with first term 6 and common difference 4. Its terms are 6, 10, 14, 18, 22, 26; therefore, 26 is the sixth term. The fifth term is 22, so it is not correct. Exam tip: count the first term as term number 1.

In an arithmetic progression, the common difference is \(d=a_2-a_1\). Therefore, \(d=19-11=8\), so \(8\) is correct. If the difference were \(9\), the second term would be \(11+9=20\), not the given term. Exam tip: when two consecutive terms are given, subtract the earlier term from the later term.

An arithmetic progression is defined by a constant difference between consecutive terms. Calculate the differences: 29 − 22 = 7, 36 − 29 = 7, and 43 − 36 = 7. Since the same difference must continue, add 7 to the last known term: 43 + 7 = 50. Thus option C is correct. Option A would add only 5, option B would add 6, and option D would add 8; none of these preserves the established common difference. The important method is to compare consecutive terms rather than estimate their size. Because every step in the given sequence increases by exactly seven, 50 is the only possible next term under the arithmetic-progression rule.

The direct answer is option C, 80. The sequence decreases by 10 each time: \(110-120=-10\), \(100-110=-10\), and \(90-100=-10\). Therefore the terms are first 120, second 110, third 100, fourth 90, and fifth \(90-10=80\). Option C is correct. Option A, 60, would be the seventh term; option B, 70, would be the sixth term; and option D, 90, is already the fourth term. Using the formula confirms it: \(a_5=120+(5-1)(-10)=120-40=80\). The minus sign is essential because the progression is decreasing. A common mistake is to move five times after starting with the first term; reaching the fifth term requires only four gaps. Memory cue: for the fifth term, use \(a_1+4d\), and here \(d=-10\).

The common difference of an arithmetic progression is found by subtracting one term from the next. The first term here is \(y\), and the second term is \(y+6\). Thus \((y+6)-y=6\), because the two occurrences of \(y\) cancel. The third term confirms the pattern: \((y+12)-(y+6)=6\).

Therefore each term is obtained by adding 6 to the preceding term, so the common difference is 6. The value of \(y\) is not needed; it may be any suitable number, because it cancels during subtraction. Option A is only the first term, and option D is the total increase across two steps, not one step. Hence option C is correct.

Consecutive terms are equal, so the difference is (8-8=0). In exams, a constant sequence can also be an arithmetic progression.

An arithmetic progression changes by the same common difference from one term to the next. If the first term is \\(a=12\\) and the common difference is \\(d=0\\), no change occurs at any step. Thus the second term is \\(12+0\\), and the third is also the preceding term plus zero. A zero difference means a constant sequence, not an alternating or increasing sequence.

Using the first-term rule, \\(a_1=12\\), \\(a_2=a_1+d=12+0=12\\), and \\(a_3=a_2+d=12+0=12\\). Therefore, the first three terms are \\(12,12,12\\), which is option C. Options A, B, and D incorrectly introduce a change between terms even though the common difference is zero.

The difference between (15) and (21) is (6), so the missing term is (15-6=9). In exams, apply the common difference backward too.

In the linear rule (a_n=7n-1), the coefficient of (n) is the difference (7). In exams, take (k) as (d) in (kn+c).

The first term of this AP is 16 and the common difference is 4. The sixth term is obtained by adding 4 five times after 16: \(a_6=16+5\times4=36\). Note that 32 is the fifth term, so it is a close but incorrect option. Exam tip: use \(a_n=a+(n-1)d\), taking care to use \(n-1\).

In an arithmetic progression, the difference between consecutive terms remains constant. In option A, 7−3=4, 11−7=4, and 15−11=4, so it is an AP. Option B has a constant ratio of 2, not a constant difference. Exam tip: check at least two consecutive differences.

This is an arithmetic progression with first term 12 and common difference 3. The seventh term is \(a_7=12+(7-1)\times3=30\). Getting 27 means adding the difference only five times. Exam tip: use \(n-1\) gaps, not \(n\).

The governing idea is finding a specified term of an arithmetic progression. Each displayed term is a consecutive multiple of 9: 9×1, 9×2, 9×3, and 9×4. Therefore the tenth term is 9×10=90. Using the standard formula confirms this result: a₁=9 and d=18−9=9, so a₁₀=a₁+(10−1)d=9+9×9=90. Hence option C is correct. Option B, 81, is the ninth multiple of 9 and therefore corresponds to the ninth term. Option A is the eighth multiple, while 99 is not a multiple of 9 and would not preserve the common difference of 9. Both the pattern and the formula lead to the same unique answer.

Given \(a_n=15-3n\), substitute \(n=4\): \(a_4=15-3(4)=15-12=3\). Therefore, 3 is correct. A value such as 6 may result from an incorrect multiplication of \(3\times4\). Exam tip: substitute the term number first, then simplify step by step.

At (n=1) it gives (12), and at (n=2) it gives (9), so (a_n=15-3n). In exams, choose the rule after checking the first term.

The middle term is the average of (8) and (20), so (x=\frac{8+20}{2}=14). In exams, take the average for the middle term in three-term progressions.

Here, the first term is \(a=20\) and the common difference is \(d=26-20=6\). The \(n\)th term is \(a_n=a+(n-1)d\). Therefore, \(a_7=20+(7-1)\times6=20+36=56\). Getting 54 is a common calculation error; the common difference must be added six times to the first term. Exam tip: for the seventh term, use \(7-1=6\) common differences.

An arithmetic progression has a fixed common difference between consecutive terms. We need a sequence that begins with 9 and increases by 5 each time. Option A is 9, 14, 19, 24, and so on. The first term is 9, and the successive differences are \(14-9=5\), \(19-14=5\), and \(24-19=5\). Therefore it meets both conditions exactly.

The other choices do not. Option B begins with 5 rather than 9. Option C begins with 9, but it adds 9 each time, since \(18-9=9\), not 5. Option D has the required difference of 5 but begins with 14, not 9. Hence option A is the only valid answer. Always check both the starting term and the change between terms, because matching only one of them is insufficient.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Therefore, \(a_6=11+(6-1)\times6=11+30=41\). Option 37 would result from adding only \(4d\), but reaching the sixth term requires adding \(5d\) to the first term. Exam tip: remember to use \(n-1\) in the nth-term formula.

The governing concept is the common difference of an arithmetic progression. In an arithmetic progression, subtracting any term from the next term gives the same constant value. Using the first two terms, d = 20 − 14 = 6. Checking the remaining consecutive pairs confirms the pattern: 26 − 20 = 6 and 32 − 26 = 6. Hence option B, 6, is correct. Option A would be obtained from an incorrect subtraction or a missed place value, while options C and D do not equal the differences between consecutive terms. The common difference is not found by adding the terms or comparing the first and last displayed terms; it is found by subtracting adjacent terms.

The governing test for an arithmetic progression is equality of consecutive differences. For the sequence shown, the first difference is 5 − 2 = 3, the second is 9 − 5 = 4, and the third is 14 − 9 = 5. Since 3, 4, and 5 are not equal, one constant common difference does not exist; therefore option A is correct. An arithmetic progression may begin with any number, including 2, so option B gives no valid reason. Its terms may increase, but increasing terms alone do not make a sequence arithmetic, so C is false. The number of displayed terms is irrelevant, making D false. The changing differences are the decisive evidence.

An arithmetic progression is identified by a constant difference between consecutive terms. In option C, 11 − 7 = 4, 15 − 11 = 4, and 19 − 15 = 4, so every consecutive difference is equal and the sequence is an arithmetic progression. Therefore option C is correct. Option A doubles each term, giving changing differences 3, 6, and 12; it is geometric rather than arithmetic. Option B consists of successive square numbers, whose differences are 5, 7, and 9, so they are not constant. Option D also doubles each term and has differences 2, 4, and 8. Equal ratios or a recognisable pattern are not enough; the defining test here is equal subtraction results.

The common difference of an arithmetic progression is the difference between consecutive terms. Here, \(9-4=5\) and \(14-9=5\), so the common difference is 5. The number 4 is the first term, not the difference. Exam tip: subtract the first term from the second term to find the common difference.

The answer is option B, 39. An arithmetic progression changes by the same common difference each time. Subtract consecutive terms: \(18-11=7\), \(25-18=7\), and \(32-25=7\). Therefore add 7 to the last term: \(32+7=39\). Option A, 37, adds only 5 to 32, so it does not continue the pattern. Option B, 39, adds the correct common difference of 7 and is correct. Option C, 41, adds 9, which is not the established difference. Option D, 43, adds 11 and is also inconsistent. The important step is to find the common difference before guessing the next term. A sequence is not continued by choosing a nearby number; its stated pattern must be preserved. Memory cue: in an AP, keep adding or subtracting the same number.

In an arithmetic progression, the second term is found by adding the common difference to the first term. Thus, second term = 9 + 4 = 13. Option 12 would require adding an incorrect difference, so it is not correct. Exam tip: To find the next term of an AP, add the common difference to the preceding term.