Arithmetic Progression

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_4=11+3\times5=26\) and \(a_{10}=11+9\times5=56\). Therefore, \(a_4+a_{10}=26+56=82\), so option B is correct. A value such as 84 can result from using one extra multiple of the common difference while finding a term. Exam tip: always use \((n-1)d\) for the \(n\)th term.

For this AP, the first term is \(a=12\) and the common difference is \(d=19-12=7\). Using \(a_n=a+(n-1)d\), \(a_{11}=12+10\times7=82\) and \(a_3=12+2\times7=26\). Therefore, \(a_{11}+a_3=82+26=108\). The nearby value 112 can result from an error in the term number or common difference. Exam tip: always use \(n-1\) in the formula for \(a_n\).

The increase over five gaps is (35), so (d=7) and (a_{13}=59+3(7)=80). First find (d), then move forward.

(a_2=a_1+5) and (a_8=a_1+35), so (2a_1+40=70) and (a_1=15). In sum-based questions, write both terms using (a_1) and (d).

In an arithmetic progression, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_{13}-a_4=(13-4)d=9d\). Here, \(73-19=54\), so \(9d=54\) and \(d=6\). Note that 9 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 first term.

Given \(a_n=9n-4\). To find the position of 113, put \(a_n=113\): \(9n-4=113\Rightarrow 9n=117\Rightarrow n=13\). Therefore, 113 is the 13th term of the progression. The 12th term is \(9(12)-4=104\), so it is not correct. Exam tip: To find the position of a given term, equate \(a_n\) to that term's value.

For three consecutive terms, (2(3x+6)=(2x+1)+(5x+5)), giving (x=6). The middle term is the average of the two surrounding terms.

In an arithmetic progression, the common difference is constant. Here, \(a_{11}-a_3=8d=72-16=56\), so \(d=7\). Therefore, \(a_7=a_3+4d=16+4\times7=44\). Option 42 is incorrect because it does not give the required difference across four terms from \(a_3\) to \(a_7\). Exam tip: When a term’s index is midway between two given indices, its value is also the average of those two terms.

In (13,18,23,28,33,38,\ldots), the sixth term is (38) and the common difference is (5). Check each option up to the required term.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Hence, \(93=9+(15-1)d=9+14d\). Therefore, \(14d=84\), so \(d=6\). If 7 were used, the 15th term would be \(9+14\times7=107\), not 93. Exam tip: there are always \(n-1\) gaps before the \(n\)th term.

The terms are (31,25,19,13,7,1,-5,\ldots), so the first negative term is the (7)th. Among the options, the (7)th is correct.

There are (9) gaps between (a_{14}) and (a_5), and (d=8), so the difference is (72). Term difference depends on position difference.

The first (6) terms are (14,19,24,29,34,39), and their sum is (159). Forming terms from the rule and adding is a safe method.

In an arithmetic progression, the difference between two terms equals the difference in their positions multiplied by the common difference. Thus, \(a_{12}-a_5=7d=68-26=42\), so \(d=6\). Now \(a_5=a_1+4d\), hence \(26=a_1+4(6)\), giving \(a_1=2\). Option 6 is the common difference, not the first term. Exam tip: use \(a_n=a_1+(n-1)d\), taking care to use \(n-1\).

(a_{10}) is equidistant from (a_6) and (a_{14}), so (a_{10}=\frac{120}{2}=60). The average of symmetric terms is the middle term.

From (2(4x+3)=(3x-2)+(6x+1)), (x=7), so the terms are (19,31,43) and the difference is (12). First find (x), then find the difference.

The terms are (22,17,12,7,2,-3,\ldots), so there are (5) positive terms. Do not count zero or negative terms as positive.

(a_{10}) is equidistant from (a_4) and (a_{16}), so (a_{10}=\frac{140}{2}=70). The average of symmetric terms is the middle term.

The direct answer is A: \((1,9,17,25,\ldots)\). For an arithmetic progression, \(a_n=a+(n-1)d\). The given terms differ by \(65-17=48\) across six index steps, from the third term to the ninth term. Hence \(6d=48\), so \(d=8\). Since \(a_3=a+2d=17\), we get \(a+16=17\), so \(a=1\). The progression is therefore \(1,9,17,25,33,41,49,57,65\). Option A has first term 1 and common difference 8, giving the required third and ninth terms. Option B has difference 7, so its ninth term is 59, not 65. Option C has difference 6, so its ninth term is 53. Option D has first term 9 and difference 8; its third term is 25, not 17. Always test both given conditions, not just one. Memory cue: the gap between the third and ninth terms contains six equal AP gaps.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Hence, \(a_5=4+4d\) and \(a_9=4+8d\). Using the given condition, \((4+4d)+(4+8d)=80\), so \(8+12d=80\). Therefore, \(12d=72\) and \(d=6\). If the common difference were 5, the sum would be \(68\), not 80. Exam tip: first express each given term in terms of \(a_1\) and \(d\).

The general term is (a_n=5n+9), and the greatest term below (120) is (119). In boundary questions, check nearby terms.

Given \(a_n=63-5n\), we get \(a_4=63-5(4)=43\) and \(a_9=63-5(9)=18\). Therefore, \(a_4+a_9=43+18=61\), so option B is correct. A value such as 66 may result from substituting the term numbers incorrectly. Exam tip: always substitute the value of \(n\) in brackets in the nth-term formula.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_{13}-a_8=5d=80-45=35\), so \(d=7\). Now, from \(a_8=a_1+7d\), we get \(45=a_1+49\), hence \(a_1=-4\). If \(-2\) were the first term, the eighth term would be \(47\), not \(45\). Exam tip: subtract the given terms first 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, \(2(2p+5)=(p+2)+(4p+1)\). This gives \(4p+10=5p+3\), so \(p=7\). If 5 is substituted, the three terms do not have a common difference. Exam tip: for three consecutive AP terms, use \(2b=a+c\) directly.

The first and ninth terms are (5) and (69), so the sum is (\frac{9(5+69)}{2}=333). Pairing symmetric terms makes calculation quick.