Arithmetic Progression

Here, the first term is \(a=13\) and the common difference is \(d=10-13=-3\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(13+(n-1)(-3)=-8\), giving \(-3(n-1)=-21\). Hence \(n-1=7\) and \(n=8\). Therefore, \(-8\) is the eighth term. The ninth term would be \(-11\), so it is not correct. Exam tip: write the negative common difference carefully before applying the \(n\)th-term formula.

(a_5=38) and (a_6=46), so the sum is (84). In exams, find both terms separately and add them.

In (S_n=\frac{n}{2}[12+5(n-1)]), putting (n=11) gives (451). In exams, you can also check quickly by substituting options.

In an arithmetic progression, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_{12}-a_5=(12-5)d\), so \(46-18=7d\). Hence, \(28=7d\) and \(d=4\). Option 7 is the difference between the term numbers, not the common difference. Exam tip: use \(a_n-a_m=(n-m)d\) to find \(d\) without first finding the initial term.

In an arithmetic progression, the third term is \(a_3=a_1+2d\). Hence, \(11=a_1+2(5)=a_1+10\), so \(a_1=1\). If 6 were the first term, the third term would be \(6+10=16\), not 11. Exam tip: use \(a_n=a_1+(n-1)d\) to relate any term to the first term.

For three consecutive terms, (2(x+4)=(2x-3)+(3x+1)), giving (x=2). The middle term is the average of the two surrounding terms.

In an arithmetic progression, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_9-a_2=7d=44-9=35\), so \(d=5\). Now, \(a_5=a_2+3d=9+3\times5=24\). Option 22 does not fit a progression with common difference 5. Exam tip: first find \(d\) from the given terms, then move to the required term.

An arithmetic progression has a constant difference, and its fourth term must be the fourth listed number. In option A, the differences are 8 − 2 = 6, 14 − 8 = 6, and 20 − 14 = 6; therefore it is an AP with d = 6, and its fourth term is 20. Option B also has common difference 6, but its fourth term is 22. Option C has fourth term 26, and option D has fourth term 38. Thus only option A satisfies both conditions simultaneously. The check must include both the required term position and the common difference, not just one of them.

Consecutive terms have equal differences, so \(y-x=z-y\). Rearranging gives \(2y=x+z\); thus, the middle term is the arithmetic mean of its neighbours. \(y=x+z\) is not generally true. In exams, use this relation to identify an AP quickly.

The terms are (18,13,8,3,-2,\ldots), so the first negative term is the (5)th. Check terms around zero in order.

There are (8) gaps between (a_{12}) and (a_4), and (d=7), so the difference is (56). Term difference depends on position difference.

The first (6) terms are (11,14,17,20,23,26), and their sum is (111). Forming terms from the rule and adding is a safe method.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_3=a_1+2d=16\) and \(a_8=a_1+7d=41\). Subtracting the equations gives \(5d=25\), so \(d=5\). Now, \(a_1=16-2\times5=6\). Therefore, 6 is correct. Option 5 is a close distractor because it is the common difference, not the first term. Exam tip: When two terms are given, subtract their equations first to find \(d\).

In an arithmetic progression, the average of (a_4) and (a_8) is (a_6), so (a_6=\frac{50}{2}=25). Equidistant terms average to the middle term.

From (2(x+2)=(x-4)+(2x-2)), (x=8), so the common difference is (10-4=6). First find (x), then find the difference.

The governing concept is the general term of an arithmetic progression: a_n = a_1 + (n - 1)d. Here a_n = 12 + (n - 1)(-3), so the first eight terms are 12, 9, 6, 3, 0, -3, -6, and -9. A positive term must be strictly greater than zero; therefore, 12, 9, 6, and 3 qualify, while 0 does not because zero is neither positive nor negative. Thus exactly 4 of the first 8 terms are positive, so option B is correct. Option A misses one positive term, whereas options C and D incorrectly include zero or a negative term.

(a_6) is equidistant from (a_2) and (a_{10}), so (a_6=\frac{64}{2}=32). The average of symmetric terms is the middle term.

For an arithmetic progression, a_n = a_1 + (n - 1)d, so the difference between a_7 and a_3 is four common-difference steps: a_7 - a_3 = 4d. The given values produce 32 - 12 = 20 = 4d, hence d = 5. Since a_3 = 12, moving backward by two steps gives a_1 = 12 - 2(5) = 2. The progression is therefore 2, 7, 12, 17, 22, 27, 32, so option B satisfies both conditions. Option A has the correct difference but its third term is 10; option C has third term 12 but difference 4; option D has third term 18. Both requirements must hold simultaneously.

Using (a_4=3+3d) and (a_6=3+5d), (6+8d=56), so (d=\frac{25}{4}). This result is not among the options.

The general term is (a_n=3n+4), and the greatest value below (100) is (97). In boundary questions check the nearest terms.