If (a_{3m}=152), (a_m=56), and (d=8), what is the value of (m)?

Correct answer: (6)

The direct answer is B: \(m=6\). In an AP, the term at position \(r\) is \(a_r=a+(r-1)d\). Instead of finding the first term, subtract the two given terms. The positions are \(3m\) and \(m\), so their difference is \(2m\). Thus \(a_{3m}-a_m=(3m-m)d=2m\times8\). The values give \(152-56=96\), so \(16m=96\), and dividing by 16 gives \(m=6\). Option A, 5, would give a difference of 80, not 96. Option B, 6, satisfies the equation. Option C, 7, would give 112, and option D, 8, would give 128; both are too large. The key idea is that AP terms change by \(d\) for every one-position increase, so multiply the index gap by \(d\).