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 has a constant common difference. Subtract consecutive terms: 29 − 22 = 7, 36 − 29 = 7, and 43 − 36 = 7. Therefore the next term is obtained by adding 7 to the last known term: 43 + 7 = 50. Thus option C is correct. Option A is too small because it adds only 5, option B adds 6, and option D adds 8. The key is to continue the same difference rather than guess from the size of the terms. Since the progression is increasing regularly by seven, 50 is the only value that preserves the arithmetic-progression rule.

The common difference is (-10), so the fifth term is (80). In exams, count in the correct direction for decreasing terms.

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 terms are consecutive multiples of 9: 9 × 1, 9 × 2, 9 × 3, and 9 × 4. Therefore the tenth term is 9 × 10 = 90. Using the arithmetic-progression formula gives the same result: a = 9, d = 18 − 9 = 9, so a_10 = 9 + (10 − 1)9 = 9 + 81 = 90. Hence option C is correct. Option B is the ninth multiple of 9, option A is the eighth multiple, and option D is not a multiple of 9 and does not continue the constant difference of 9.

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.