For which (p) will (4,4+p,4+3p,4+6p,\ldots) be an arithmetic progression?

Correct answer: Only (p=0)

The direct answer is option A: only
\(p=0\). An arithmetic progression has the same difference between every pair of consecutive terms. Here the first differences are
\((4+p)-4=p\), then
\((4+3p)-(4+p)=2p\), and then
\((4+6p)-(4+3p)=3p\). For an AP, we need
\(p=2p=3p\). The first equality gives
\(p=0\), and this value also makes every difference zero, so the sequence becomes
\(4,4,4,4,\ldots\). Option A is correct. Option B,
\(p=1\), gives differences
\(1,2,3\), which are unequal. Option C is wrong because an arbitrary p does not make these differences equal. Option D is wrong because p=0 does work. The useful method is always to subtract consecutive terms before choosing an answer.