Arithmetic Progression

To find the zero term, set the given term equal to 0: \(24-4n=0\). Thus, \(4n=24\), so \(n=6\). Therefore, the sixth term of the progression is zero. For the fifth term, \(24-4(5)=4\), so it is not zero. Exam tip: To find the position of a specified term, substitute its value for \(a_n\) and solve for \(n\).

The increase over eight gaps is (48), so (d=6) and (a_{10}=31+4(6)=55). (a_{10}) lies between the two given terms.

In (36,32,28,24,20,16,12,8,4,\ldots), the first term is (36) and the ninth term is (4). Check options up to the ninth term.

There are (9) gaps between (a_{16}) and (a_7), so the difference is (9\times9=81). Term differences can be found directly using (d).

The general term is (a_n=6n+11), and (a_{16}=107) is the first term greater than (100). In boundary questions, check nearby terms.

(a_{10}) is equidistant from (a_5) and (a_{15}), so (a_{10}=\frac{150}{2}=75). The average of symmetric terms is the middle term.

For this AP, the first term is \(a=7\) and the common difference is \(d=9\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(a_6=7+5\times9=52\) and \(a_{10}=7+9\times9=88\). Therefore, \(a_6+a_{10}=52+88=140\). A value such as 138 can result from an off-by-one error while counting terms, but the correct sum is 140. Exam tip: always use \((n-1)d\) for the \(n\)th term.

For an arithmetic progression, \(a_n=a+(n-1)d\). Therefore, \(a_{16}-a_7=(16-7)d=9d\). Given \(96-33=63\), we get \(9d=63\), so \(d=7\). Do not confuse the 9 gaps with the value of the common difference. Exam tip: when two terms are given, use \(a_m-a_n=(m-n)d\).

For three consecutive terms of an arithmetic progression, the middle term is the average of the first and third terms. Hence, \(2(6x+1)=(4x-3)+(9x-2)\). This gives \(12x+2=13x-5\), so \(x=7\). Therefore, option C is correct. For example, if \(x=6\), the two consecutive differences are not equal. Exam tip: for three consecutive AP terms, directly use \(2b=a+c\).

The direct answer is B: 5. For an arithmetic progression, the sum of the first n terms is S_n = n/2[2a+(n-1)d]. Here n=7, a=8 and S_7=161. Substitute: 161 = 7/2[2(8)+6d] = 7/2(16+6d). Multiplying by 2 gives 322 = 7(16+6d), so 322 = 112+42d, 210=42d and d=5. Equivalently, the average of seven terms is 161/7=23; the middle fourth term is 23, and 8+3d=23, giving d=5. Option A, 4, would give a sum of 140. Option B is correct. Option C, 6, would give a sum of 182. Option D, 7, would give 203. The common mistake is forgetting n-1 and writing 7d instead of 6d in the formula.

The increase over four gaps is (24), so (d=6) and (a_{12}=44+4(6)=68). Moving forward from a nearby term is easy.

The general term is (a_n=6n+3), and the greatest term below (90) is (87). In boundary questions, check nearby terms.

In an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_4=45+3(-6)=27\) and \(a_8=45+7(-6)=3\). Therefore, \(a_4+a_8=27+3=30\), so option B is correct. A value such as \(36\) can result from using the wrong number of common differences. Exam tip: for the \(n\)th term, use \(n-1\) common differences.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_{14}-a_5=9d=81-18=63\), so \(d=7\). Now, from \(a_5=a_1+4d\), we get \(18=a_1+28\), hence \(a_1=-10\). If \(-8\) were used, the fifth term would be \(20\), so it is not correct. Exam tip: subtract the given terms first to find \(d\) quickly.

The first and twelfth terms are (7) and (84), so the sum is (\frac{12(7+84)}{2}=546). Use the first and last terms for the sum.

The direct answer is option C, the 8th term. The formula is a_n=39-5n, so substitute each relevant counting number. For n=6, a_6=39-30=9, which is positive. For n=7, a_7=39-35=4, also positive. For n=8, a_8=39-40=-1, which is negative. Therefore the first negative term is the 8th term; earlier terms are still above zero. Option A, the 6th term, is wrong because its value is 9. Option B, the 7th term, is wrong because its value is 4. Option C works because its value is -1. Option D, the 9th term, is also negative, but it is not the first negative term: a_9=39-45=-6. The word “first” means that all earlier terms must be checked. Memory cue: continue until the value changes from positive to negative.

(a_7) is equidistant from (a_4) and (a_{10}), so (a_7=\frac{96}{2}=48). The average of symmetric terms is the middle term.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_5=14+4\times6=38\) and \(a_{11}=14+10\times6=74\). Therefore, \(a_5+a_{11}=38+74=112\). A value such as 116 can result from adding the common difference an incorrect number of times. Exam tip: always use \((n-1)d\) for the \(n\)th term.

The first term of this AP is 16 and the common difference is 7. Thus, \(a_{12}=16+11\times7=93\) and \(a_4=16+3\times7=37\). Therefore, \(a_{12}+a_4=93+37=130\), so option C is correct. A value such as 123 can result from counting the number of common differences incorrectly. Exam tip: always use \(n-1\) in \(a_n=a+(n-1)d\).

The increase over five gaps is (35), so (d=7) and (a_{15}=64+4(7)=92). First find (d), then move forward.

(a_3=a_1+12) and (a_{11}=a_1+60), so (2a_1+72=92) and (a_1=10). In sum-based questions, write both terms using (a_1) and (d).