Arithmetic Progression

In an arithmetic progression, the difference between consecutive terms must remain constant. Here, 9−5=4, 14−9=5, and 18−14=4, so it is not an AP. Increasing terms alone are not enough. In exams, always check consecutive differences first.

The first term is (8) and the difference is (6), so (a_n=8+(n-1)6=6n+2). In the general term, (d) appears with (n).

In an arithmetic progression, the fourth term is \(a_4=a_1+3d\), because there are three equal gaps between the first and fourth terms. Thus, \(27=12+3d\), so \(3d=15\) and \(d=5\). If the difference were 6, the fourth term would be \(12+3\times6=30\), not 27. Exam tip: use \(a_n=a+(n-1)d\) for the nth term.

The first term is (4), and adding (7) each time gives (4,11,18,25,\ldots). Both the first term and the difference must match.

The first (6) terms are (4,9,14,19,24,29), and their sum is (99). For small terms, listing and adding works well.

In this arithmetic progression, the first term is \(a=1\) and the common difference is \(d=6-1=5\). The \(n\)th term is given by \(a_n=a+(n-1)d\). Therefore, \(a_{12}=1+(12-1)\times5=1+55=56\). Hence, 56 is correct. The option 51 results from the common mistake of adding only 10 differences instead of 11. Exam tip: always use \((n-1)\) when finding the \(n\)th term of an AP.

In an arithmetic progression, the second term is \(a_2=a_1+d\). Hence, \(a_1=a_2-d=13-4=9\). Therefore, the correct answer is 9. Choosing 8 would result from an incorrect subtraction. Exam tip: To find the first term from the next term, subtract the common difference.

In (3,6,12,24,\ldots), the differences are (3,6,12), which are not equal. For an arithmetic progression, the difference must always be constant.

(a_n=40+(n-1)(-5)), and (40-5(n-1)=5) gives (n=8). In a decreasing progression, the position is still positive.

In an arithmetic progression, the seventh term is \(a_7=a_1+6d\) because there are 6 gaps between the first and seventh terms. Thus, \(33=3+6d\), so \(6d=30\) and \(d=5\). If the common difference were 6, the seventh term would be \(3+6\times6=39\), which is incorrect. Exam tip: In \(a_n=a+(n-1)d\), remember to use \(n-1\), not \(n\).

Given \(a_n=4n-1\). To find the eighth term, substitute \(n=8\): \(a_8=4(8)-1=32-1=31\). Hence, 31 is correct. The value 29 could result from an incorrect multiplication. Exam tip: after substituting the value of \(n\), check multiplication and subtraction separately.

The first four terms of the given arithmetic progression are 7, 10, 13, and 16. Therefore, their sum is \(7+10+13+16=46\). Hence, option C is correct. The value 44 is the sum of only the first three terms, so it is not correct. Exam tip: When the number of terms is small, direct addition is usually the quickest method.

In (18,12,6,0,\ldots), (6) is subtracted each time, so (d=-6). The sign is very important in common difference.

From (a_4) to (a_9), there are (5) gaps, so (a_9=18+5(3)=33). Moving forward from a nearby term is a quick method.

To find the first term, substitute \(n=1\). Thus, \(a_1=9(1)+2=11\), so 11 is correct. Option 9 is only the coefficient of \(9n\); the constant term 2 must also be added. Exam tip: To obtain the first term from any \(a_n\), always put \(n=1\).

The first term is (5), and adding (2) each time gives (5,7,9,11). Form terms directly from the given (a) and (d).

From the third to the fifth term there are (2) gaps and the increase is (8), so (d=4). Count the number of gaps between terms.

The first term is (22) and (d=-4), so (a_n=22+(n-1)(-4)=26-4n). Keep (d) negative in a decreasing sequence.

The average of the first three terms is (\frac{2+8+14}{3}=8). In an arithmetic progression, the average of three consecutive terms is the middle term.

This is an AP with \(a=20\), \(d=3\), and \(n=10\). So, \(S_{10}=\frac{10}{2}[2(20)+9(3)]=5(67)=335\). Finding only the 10th term does not give the total seats. Exam tip: use \(S_n\) when a total is asked.