Arithmetic Progression

In an arithmetic progression, \(a_n=a+(n-1)d\). Therefore, \(a_8=a+7d\). Substituting the given values gives \(66=17+7d\), so \(7d=49\) and \(d=7\). Hence, option C is correct. If \(d=8\), the eighth term would be \(17+7\times8=73\), not 66. Exam tip: use \(n-1\), not \(n\), in the formula for the \(n\)th term.

Here, the first term is \(a=24\) and the common difference is \(d=19-24=-5\). Using \(a_n=a+(n-1)d\), we get \(24+(n-1)(-5)=-11\). Thus, \(29-5n=-11\), which gives \(n=8\). Therefore, \((-11)\) is the eighth term. The seventh term is \((-6)\), so it is a close but incorrect option. Exam tip: In a decreasing AP, make sure to use a negative common difference.

The nth term of an arithmetic progression is \(a_n=a+(n-1)d\). Thus, \(a_5=12+(5-1)\times9=48\) and \(a_8=12+(8-1)\times9=75\). Therefore, \(a_5+a_8=48+75=123\). A value such as 117 may result from counting the common difference incorrectly. Exam tip: write \(n-1\) first while finding each term, then add the terms.

In (S_n=\frac{n}{2}[14+6(n-1)]), (n=11) gives (407) and (n=12) gives (480), so no option gives (539). In exams, check by substituting options.

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.

The governing formula is a_n = a_1 + (n - 1)d. For the fifth term, a_5 = a_1 + 4d. Substituting the given values gives 28 = a_1 + 4(7) = a_1 + 28. Therefore a_1 = 28 - 28 = 0, so option A is correct. The result can also be understood by moving backward four equal differences from the fifth term: 28, 21, 14, 7, 0. A common error is to subtract only three differences or to use a_5 = a_1 + 5d; the expression uses n - 1 differences because the first term is the starting point. The other options therefore do not reproduce the given fifth term with d = 7.

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.

In (1,9,17,25,\ldots), (d=8), so (a_3=17) and (a_9=65). Check both conditions in the options.