Arithmetic Progression

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.

(a_7=4) and (a_8=-1), so the first negative term is the (8)th. Look for the first term below zero.

(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).

The governing concept is the nth-term formula for an arithmetic progression, aₙ = a₁ + (n − 1)d. Since a₁₀ − a₄ spans six equal steps, 57 − 21 = 36 = 6d, so the common difference is d = 6. Moving backward three steps from a₄ to a₁ gives a₁ = 21 − 3(6) = 3. Therefore aₙ = 3 + (n − 1)6 = 6n − 3. Direct checks give a₄ = 24 − 3 = 21 and a₁₀ = 60 − 3 = 57. Option B has the wrong first term, option C has the wrong common difference, and option D gives a₄ = 21 but fails at a₁₀.