Arithmetic Progression

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.

The answer is option C, the 9th term. We need the first term for which \(a_n<0\). Given \(a_n=25-3n\), test the boundary: \(25-3n<0\), so \(25<3n\), hence \(n>25/3=8\frac{1}{3}\). The smallest whole-number position is \(n=9\). Direct checking gives \(a_8=25-24=1\), still positive, and \(a_9=25-27=-2\), negative. Option A, 7th, gives 4, positive. Option B, 8th, gives 1, positive. Option C, 9th, gives -2 and is the first negative term. Option D, 10th, gives -5, which is negative but not the first one. The key is to use a strict inequality because “negative” means less than zero, not equal to zero. Memory cue: find the first integer after the expression crosses below zero.

(a_4) is equidistant from (a_2) and (a_6), so (a_4=\frac{40}{2}=20). The average of symmetric terms is the middle term.

The nth term of an arithmetic progression is \(a_n=a_1+(n-1)d\). Thus, \(a_3=9+(3-1)\times4=17\) and \(a_9=9+(9-1)\times4=41\). Therefore, \(a_3+a_9=17+41=58\), so option C is correct. Choosing 56 may result from an error in finding the term number or multiplying the common difference. Exam tip: always add \((n-1)d\) to the first term when finding the nth term.

In this arithmetic progression, the first term is 8 and the common difference is 7. Thus, \(a_{10}=8+(10-1)\times7=71\) and \(a_2=8+(2-1)\times7=15\). Hence, \(a_{10}+a_2=71+15=86\). A value such as 84 may result from using an incorrect common difference or term position. Exam tip: always use \(n-1\) in \(a_n=a+(n-1)d\).

The increase over five gaps is (30), so (d=6) and (a_{12}=48+3(6)=66). First find (d), then move forward.

For an arithmetic progression, \(a_{14}-a_6=(14-6)d\). Thus, \(61-21=8d\), so \(40=8d\) and hence \(d=5\). If \(d=4\), the difference across 8 term positions would be \(32\), not \(40\). Exam tip: When subtracting two AP terms, also subtract their term numbers.

The nth term is \(a_n=8n-5\). To find the position of 83, set \(8n-5=83\). This gives \(8n=88\), so \(n=11\). Hence, 83 is the 11th term. The 10th term is \(75\), so it is a close but incorrect option. Exam tip: to find the position of a given term, equate \(a_n\) to that value and solve for \(n\).

For three consecutive terms, (2(2x+1)=(x-3)+(4x-1)), giving (x=6). The middle term is the average of the two surrounding terms.

In an AP, a fixed common difference d is added to each term, so aₙ₊₁−aₙ=d. A constant ratio identifies a geometric progression, not necessarily an AP. Exam tip: compare consecutive differences to identify an AP.

In (5,11,17,23,29,\ldots), the fifth term is (29) and the common difference is (6). Check each option up to the required term.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(73=7+(12-1)d=7+11d\). Hence \(11d=66\), so \(d=6\). There are 11 gaps from the 1st term to the 12th term, not 12. Exam tip: always use \(n-1\) in the formula for the \(n\)th term.

The terms are (25,19,13,7,1,-5,\ldots), so the first negative term is the (6)th. Check terms around zero in order.

There are (9) gaps between (a_{15}) and (a_6), and (d=7), so the difference is (63). Term difference depends on position difference.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_4=a_1+3d=22\) and \(a_9=a_1+8d=52\). Subtracting the equations gives \(5d=30\), so \(d=6\). Now, \(a_1=22-3\times6=4\). Therefore, the correct answer is 4. Option 6 is the common difference \(d\), not the first term. Exam tip: When two terms are given, subtract their equations first to find \(d\).

(a_9) is equidistant from (a_5) and (a_{13}), so (a_9=\frac{86}{2}=43). The average of symmetric terms is the middle term.

From (2(3x+2)=(2x-1)+(5x+1)), (x=4), so the common difference is (14-7=7). Calculate (x) first and then the difference.

The terms are (18,14,10,6,2,-2,\ldots), so there are (5) positive terms. Do not count zero or negative terms as positive.

(a_9) is equidistant from (a_3) and (a_{15}), so (a_9=\frac{108}{2}=54). The average of symmetric terms is the middle term.