Arithmetic Progression

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.

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 \(2b=a+c\).

(\frac{7}{2}[2(8)+6d]=161) gives (56+21d=161) and (d=5). In sum questions, substitute (n) and (a) first.

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.