Arithmetic Progression

In (2,9,16,23,\ldots), (a_4) is not (19); the correct sequence would be (-2,5,12,19,\ldots). None of the given options is correct.

(a_4=6+3d) and (a_8=6+7d), so (12+10d=84), giving (d=7.2), which is not in the options. Option consistency must be checked.

The general term is (a_n=5n+6), and the greatest term below (90) is (86). In boundary questions, check nearby terms.

Given \(a_n=52-4n\), \(a_4=52-4(4)=36\) and \(a_9=52-4(9)=16\). Therefore, \(a_4+a_9=36+16=52\). The option 48 may result from substituting an incorrect value of \(n\) in a term. Exam tip: substitute the term number carefully into the formula before simplifying.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_{12}-a_7=5d=69-34=35\), so \(d=7\). Now, using \(a_7=a_1+6d\), we get \(34=a_1+42\), hence \(a_1=-8\). If \(-6\) were chosen, the seventh term would be \(36\), not the given \(34\). Exam tip: When two terms are given, first use the difference in their term numbers to find \(d\).

From (2(p+10)=p+(3p-2)), (2p+20=4p-2), so (p=11). The middle term is the average of the two surrounding terms.

The first (8) terms go from (4) to (60), so the sum is (\frac{8(4+60)}{2}=256). Pairing symmetric terms makes calculation quick.

To find the zero term, set the given term equal to 0: \(18-3n=0\). Thus, \(3n=18\), so \(n=6\). Therefore, the sixth term is zero. For example, the fifth term is \(a_5=18-15=3\), 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 (36), so (d=6) and (a_8=14+3(6)=32). (a_8) lies between the two given terms.

In (24,20,16,12,8,4,0,\ldots), the first term is (24) and the seventh term is (0). Check options up to the seventh term.

There are (9) gaps between (a_{14}) and (a_5), so the difference is (9\times8=72). Term differences can be found directly using (d).

The general term is (a_n=7n+6), and (a_{14}=104) is the first term greater than (100). In boundary questions, check nearby terms.

(a_8) is equidistant from (a_4) and (a_{12}), so (a_8=\frac{112}{2}=56). The average of symmetric terms is the middle term.

The first term is \(a=5\) and the common difference is \(d=12-5=7\). Using \(a_n=a+(n-1)d\), we get \(a_5=5+4\times7=33\) and \(a_9=5+8\times7=61\). Therefore, \(a_5+a_9=33+61=94\). A nearby option such as 98 may result from using an incorrect term number or common difference. Exam tip: first identify \(a\) and \(d\), then apply the formula for \(a_n\).

In an arithmetic progression, the difference between two terms equals the difference of their indices multiplied by the common difference. Thus, \(a_{15}-a_6=(15-6)d=9d\). Hence, \(82-28=54=9d\), giving \(d=6\). Option 9 is the difference between the term numbers, not the common difference. Exam tip: use \(a_m-a_n=(m-n)d\) to find \(d\) without first finding the first term.

For three consecutive terms of an arithmetic progression, twice the middle term equals the sum of the first and third terms. Thus, \(2(5x-2)=(3x+1)+(8x-8)\). This gives \(10x-4=11x-7\), so \(x=3\). On checking, the terms are \(10,13,16\), with common difference \(3\). Exam tip: for three consecutive AP terms, use \(2b=a+c\) directly.

Direct answer: Option C, 6. For an arithmetic progression, the sum of the first n terms is \\(S_n=\\frac n2[2a+(n-1)d]\\). Here \\(a=5\\), \\(n=6\\), and \\(S_6=120\\). Substitute carefully: \\(120=\\frac62[2(5)+(6-1)d]\\). Thus \\(120=3(10+5d)=30+15d\\). Subtract 30: \\(90=15d\\). Divide by 15: \\(d=6\\). Option A, 4, would give sum \\(3(10+20)=90\\), not 120. Option B, 5, would give \\(3(10+25)=105\\). Option C, 6, gives \\(3(10+30)=120\\), so it is correct. Option D, 7, would give \\(3(10+35)=135\\). A direct check gives terms 5, 11, 17, 23, 29, 35, whose sum is 120. Remember: in the sum formula use n−1, not n, and solve the equation step by step.

In an arithmetic progression, subtracting a term from the next term gives one fixed value, called the common difference. The first term is only the starting term. Exam tip: subtract consecutive terms to identify an AP.

The increase over four gaps is (24), so (d=6) and (a_{11}=36+4(6)=60). Moving forward from a nearby term is easy.

The general term is (a_n=6n+2), and the greatest term below (80) is (74). In boundary questions, check nearby terms.

The nth term of an arithmetic progression is \(a_n=a_1+(n-1)d\). Thus, \(a_3=30+2(-5)=20\) and \(a_7=30+6(-5)=0\). Therefore, \(a_3+a_7=20+0=20\). Choosing 25 can result from incorrectly calculating the number of common differences. Exam tip: always use \((n-1)d\) while finding \(a_n\).

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_{13}-a_4=9d=67-13=54\), so \(d=6\). Now, using \(a_4=a_1+3d\), we get \(13=a_1+18\), hence \(a_1=-5\). If \(-4\) were the first term, the fourth term would be \(14\), so it is not correct. Exam tip: subtracting two given terms is a quick way to find the common difference \(d\).

The first and tenth terms are (3) and (57), so the sum is (\frac{10(3+57)}{2}=300). Use the first and last terms for the sum.

The direct answer is C: the 8th term. A sequence is negative when its value is less than zero. Here the nth term is given by \\(a_n=31-4n\\). Test the terms in order: \\(a_6=31-4(6)=31-24=7\\), so the 6th term is positive; \\(a_7=31-28=3\\), so the 7th term is still positive; \\(a_8=31-32=-1\\), so the 8th term is negative. Thus the first negative term is the 8th term. A is wrong because the 6th term is 7. B is wrong because the 7th term is 3. C is correct because its value is -1. D is not the first negative term; although \\(a_9=-5\\) is also negative, negativity already began at term 8. Exam cue: calculate consecutive terms until the first value below zero.

(a_5) is equidistant from (a_3) and (a_7), so (a_5=\frac{60}{2}=30). The average of symmetric terms is the middle term.