Arithmetic Progression

(a_8=1) and (a_9=-2), so the first negative term is the (9)th. Look for the first term below zero.

(a_4) is equidistant from (a_2) and (a_6), so (a_4=\frac{40}{2}=20). The average of symmetric terms is the middle term.

The governing property is that consecutive terms of an arithmetic progression have a constant difference. For three listed terms p, q, r to be consecutive AP terms, q - p must equal r - q, or equivalently 2q = p + r. In option B, 12 - 6 = 6 and 18 - 12 = 6, so the differences are equal and the terms can be consecutive AP terms. Option A has differences 4 and 5; option C has differences 5 and 8; and option D has differences 6 and 9. Since none of those pairs is equal, those options cannot represent three consecutive terms of an arithmetic progression.

The nth term of an arithmetic progression is \(a_n=a_1+(n-1)d\). Thus, \(a_3=9+(3-1)\times4=17\) and \(a_9=9+(9-1)\times4=41\). Therefore, \(a_3+a_9=17+41=58\), so option C is correct. Choosing 56 may result from an error in finding the term number or multiplying the common difference. Exam tip: always add \((n-1)d\) to the first term when finding the nth term.

Use the arithmetic-progression formula a_n = a_1 + (n - 1)d. For the fifth term, 30 = a_1 + (5 - 1)7 = a_1 + 28. Subtracting 28 from both sides gives a_1 = 30 - 28 = 2. Therefore option A is correct. The same result can be understood by moving backward four equal steps from the fifth term: 30, 23, 16, 9, 2. Option B would result from subtracting only three differences, which reaches the second term rather than the first; options C and D similarly fail to move back the required four steps. The index difference between the first and fifth terms is four, not five.

In this arithmetic progression, the first term is 8 and the common difference is 7. Thus, \(a_{10}=8+(10-1)\times7=71\) and \(a_2=8+(2-1)\times7=15\). Hence, \(a_{10}+a_2=71+15=86\). A value such as 84 may result from using an incorrect common difference or term position. Exam tip: always use \(n-1\) in \(a_n=a+(n-1)d\).

The increase over five gaps is (30), so (d=6) and (a_{12}=48+3(6)=66). First find (d), then move forward.

Consecutive AP terms have common difference \(d\). Thus \(q=p+d\) and \(r=p+2d\), so \(p+r=2p+2d=2q\). Option A is not generally true. Exam tip: the middle term is the average of the end terms.

Here (d=-8), so (a_{10}=96+9(-8)=24). In exams, keep (d) negative in a decreasing progression.

To find which term equals 121, equate the general term to 121: \(7n-5=121\). Thus, \(7n=126\), so \(n=18\). Therefore, 121 is the 18th term of the sequence. The 17th term is \(114\), so it is not correct. Exam tip: To find the position of a given value, set \(a_n\) equal to that value.

For an arithmetic progression, the nth-term formula is \(a_n=a+(n-1)d\). Here, \(93=18+(16-1)d\), so \(93=18+15d\). Thus, \(15d=75\) and \(d=5\). The value 6 can result from incorrectly counting the number of gaps between the first and sixteenth terms. Exam tip: there are always \(n-1\) gaps from the first term to the nth term.

The middle term is (x=\frac{10+34}{2}=22). In exams, the middle term of three AP terms is the average.

In an arithmetic progression, the fourth term is \(a_4=a+(4-1)d=a+3d\). Thus, \(29=a+3(6)=a+18\), so \(a=11\). If 13 were chosen, the fourth term would be \(13+18=31\), not the given 29. Exam tip: use \(a_n=a+(n-1)d\) for the \(n\)th term.

The rule (a_n=36-5n) gives (31) at (n=1) and (26) at (n=2). In exams, match the first term in a decreasing progression.

The formula for an arithmetic progression is \(a_n=a+(n-1)d\). Hence, \(a_7=12+(7-1)d=12+6d\). So, \(54=12+6d\), giving \(6d=42\) and \(d=7\). If \(d=6\), the seventh term would be \(12+6\times6=48\), not 54. Exam tip: use \(n-1\), not \(n\), in the formula for the \(n\)th term.

Here, the first term is \(a=3\) and the common difference is \(d=8\). The \(n\)th term is \(a_n=a+(n-1)d\). Thus, \(3+(n-1)\times8=83\), giving \((n-1)\times8=80\), so \(n=11\). Therefore, 83 is the eleventh term. The twelfth term would be \(91\), so that option is not correct. Exam tip: To find the position of a number in an AP, equate it to \(a_n\) and solve for \(n\).

The nth term of an arithmetic progression is given by \(a_n=a+(n-1)d\). Therefore, \(a_8=17+(8-1)(-4)=17-28=-11\). Hence, option C is correct. A value such as \(-9\) can result from using the wrong term number or forgetting the \((n-1)\) factor. Exam tip: when \(d\) is negative, each successive term decreases.

In an AP, consecutive differences are equal: \(q-p=r-q\). Rearranging gives \(2q=p+r\), so the middle term is the average of the end terms. \(q^2=pr\) is associated with a GP. Exam tip: use equal differences to test an AP quickly.

Here, the first term is \(a=2\), the common difference is \(d=4\), and \(n=18\). The 18th term is \(a_{18}=a+17d=2+17\times4=70\). Therefore, \(S_{18}=\frac{18}{2}(a+a_{18})=9(2+70)=648\). Hence, 648 is correct. An option such as 638 can result from a small error in finding the last term or multiplication. Exam tip: for the \(n\)th term, use \((n-1)d\), not \(nd\).

(a_9-a_5=4d=28), so (d=7). In exams, divide the difference of terms by the difference of positions.