Arithmetic Progression

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 \(2\times\text{middle term}=\text{first term}+\text{third term}\).

The governing concept is the sum formula for the first n terms of an arithmetic progression: S_n = n/2[2a_1 + (n - 1)d]. Here a_1 = 2, d = 8 - 2 = 6, and n = 7. Thus S_7 = 7/2[2(2) + 6(6)] = 7/2(4 + 36) = 7/2 x 40 = 140. Listing the terms gives 2, 8, 14, 20, 26, 32, 38, whose total is also 140. Therefore option C is correct. Pairing the first and last terms gives 2 + 38 = 40, and every symmetric pair has this total; the distractors result from an arithmetic addition or term-counting error.

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.

For three consecutive terms of an arithmetic progression, the middle term is the arithmetic mean of the first and third. Therefore 2(5x + 2) = (4x - 1) + (7x - 3). Expanding gives 10x + 4 = 11x - 4, so x = 8. Substitution verifies the terms: 4(8) - 1 = 31, 5(8) + 2 = 42, and 7(8) - 3 = 53; the two successive differences are 11 and 11. Hence option D is correct. The original option set did not contain 8 and its marked answer was inconsistent, so the final option set has been corrected. Values 4, 5, and 6 fail the equal-difference condition.

The first (6) terms sum to (72) for (2,6,10,14,18,22), so (d=4). For small (n), option checking is quick.

Here, the first term is 12 and the common difference is 5. Therefore, the nth term is \(a_n=12+(n-1)\times5=5n+7\). Substituting \(a_n=72\), we get \(5n+7=72\), so \(5n=65\) and \(n=13\). Thus, 72 is the 13th term of the progression. The nearby option 14 is incorrect because the 14th term is \(77\). Exam tip: To find a term position, first use \(a_n=a+(n-1)d\).

The increase over three gaps is (12), so (d=4) and (a_9=20+4(4)=36). Moving forward from a nearby term is easy.

The general term is (a_n=4n+2), and the greatest term below (50) is (46). In boundary questions check nearby terms.

In an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_2=20+(2-1)(-4)=16\) and \(a_6=20+(6-1)(-4)=0\). Therefore, \(a_2+a_6=16+0=16\), so option A is correct. The value 20 may result from not applying the common difference \(-4\) correctly to the terms. Exam tip: Write \((n-1)d\) before substituting values for any term.

In an arithmetic progression, the difference between two terms equals the difference of their positions multiplied by the common difference. Thus, \(a_{12}-a_3=9d=61-7=54\), so \(d=6\). Now, from \(a_3=a_1+2d\), we get \(7=a_1+12\), hence \(a_1=-5\). If \(-4\) were used, the third term would be \(8\), not \(7\). Exam tip: first find \(d\) from the two given terms, then use \(a_n=a_1+(n-1)d\).

The first and ninth terms are (1) and (41), so the sum is (\frac{9(1+41)}{2}=189). Use the first and last terms for the sum.