Arithmetic Progression

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.

(\frac{6}{2}[2(5)+5d]=120) gives (30+15d=120) and (d=6). In sum questions, substitute (n) and (a) first.

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.

(a_7=3) and (a_8=-1), so the first negative term is the (8)th. Look for the first term 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.

For an arithmetic progression, \(a_n=a_1+(n-1)d\). Thus, \(a_4=11+3\times5=26\) and \(a_{10}=11+9\times5=56\). Therefore, \(a_4+a_{10}=26+56=82\), so option B is correct. A value such as 84 can result from using one extra multiple of the common difference while finding a term. Exam tip: always use \((n-1)d\) for the \(n\)th term.

Use the arithmetic-progression formula aₙ = a₁ + (n − 1)d. For n = 6, the sixth term is a₆ = a₁ + 5d because five common differences separate the first and sixth terms. Substituting the data gives 41 = a₁ + 5(8) = a₁ + 40. Therefore, a₁ = 41 − 40 = 1, so option A is correct. A backward check gives the same result: 41, 33, 25, 17, 9, 1 after subtracting 8 five times. Options B, C, and D would produce sixth terms 43, 45, and 47 respectively, so they cannot satisfy the given condition.

For this AP, the first term is \(a=12\) and the common difference is \(d=19-12=7\). Using \(a_n=a+(n-1)d\), \(a_{11}=12+10\times7=82\) and \(a_3=12+2\times7=26\). Therefore, \(a_{11}+a_3=82+26=108\). The nearby value 112 can result from an error in the term number or common difference. Exam tip: always use \(n-1\) in the formula for \(a_n\).

The increase over five gaps is (35), so (d=7) and (a_{13}=59+3(7)=80). First find (d), then move forward.

(a_2=a_1+5) and (a_8=a_1+35), so (2a_1+40=70) and (a_1=15). In sum-based questions, write both terms using (a_1) and (d).

Here (a=16) and (d=7), so (a_{13}=16+12\times7=100). In exams, use (a+(n-1)d) for the (n)th term.

In an AP, the difference between every pair of consecutive terms is constant, so \(q-p\), \(r-q\), and \(s-r\) must be equal. \(p+s=q+r\) alone is not sufficient. Exam tip: compare all three differences.

Here (d=-9), so (a_9=120+8(-9)=48). In exams, keep (d) negative in a decreasing progression.