Arithmetic Progression

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.

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.

(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.

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.

Given \(a_n=40-5n\), we get \(a_3=40-5(3)=25\) and \(a_7=40-5(7)=5\). Therefore, \(a_3+a_7=25+5=30\). The value 25 is only \(a_3\), not the required sum. Exam tip: Substitute each required value of \(n\) separately before adding the terms.

Use the difference between the two given terms:

a_{11}-a_6=(a_1+10d)-(a_1+5d)=5d.

Thus,

5d=55-25=30, so d=6. Now, from

a_6=a_1+5d, we get a_1=25-5(6)=-5. Hence, option A is correct. If a_1 were -4, the sixth term would be 26, not 25. Exam tip: when two AP terms are given, first use the difference in their term numbers to find d.

For three consecutive terms of an arithmetic progression, twice the middle term equals the sum of the first and third terms. Thus, \(2(a+8)=a+(3a-4)\). This gives \(2a+16=4a-4\), so \(2a=20\) and hence \(a=10\). Therefore, option B is correct. If \(a=12\), the differences between the three terms are not equal. Exam tip: for three consecutive AP terms, use \(2\times\text{middle term}=\text{first term}+\text{third term}\).

To find the zero term, set the given term equal to 0: \(12-2n=0\). Thus, \(2n=12\), so \(n=6\). Hence, the sixth term of the AP is zero. At the fifth term, \(12-2(5)=2\), so it is not zero. Exam tip: When a term with a specified value is asked, equate \(a_n\) to that value and solve for \(n\).

The increase over six gaps is (24), so (d=4) and (a_7=10+3(4)=22). (a_7) lies between the two given terms.

In (16,12,8,4,0,\ldots), the first term is (16) and the fifth term is (0). Check options up to the fifth term.

There are (6) gaps between (a_{15}) and (a_9), so the difference is (6\times9=54). Term differences can be found directly using (d).

The general term is (a_n=7n+2), and (a_{15}=107) is the first term greater than (100). In boundary questions check nearby terms.

(a_6) is equidistant from (a_3) and (a_9), so (a_6=\frac{72}{2}=36). The average of symmetric terms is the middle term.

Here, the first term is \(a=3\) and the common difference is \(d=7\). Using \(a_n=a+(n-1)d\), we get \(a_4=3+3\times7=24\) and \(a_8=3+7\times7=52\). Therefore, \(a_4+a_8=24+52=76\). The value 74 can result from not applying the common difference of 7 correctly. Exam tip: when finding the \(n\)th term of an AP, use \((n-1)d\), not \(nd\).

In an arithmetic progression, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_{13}-a_5=(13-5)d=8d\). Hence, \(65-17=48=8d\), giving \(d=6\). Option 8 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.