How TDM works. A commutator visits the input channels in a fixed cyclic order and interleaves their bits onto a single high-rate line. The number of slots a channel receives per frame is proportional to its bit rate, so the composite (line) rate is the sum of the channel rates. Each statement pairs a bit sequence with a rate condition, and both halves must agree.

A — "A₁B₁C₁D₁, all channels at the same bit rate." CONSISTENT. Every channel gets exactly one slot per frame, so all four are sampled equally often and share the line rate equally:

\(R_{line} = 4R_{channel}\)

This is ordinary equal-rate TDM.

C — "A₁B₁A₂C₁A₃D₁, channel A at three times the rate of B, C, D." CONSISTENT. Count the slots in one frame: A appears three times (A₁, A₂, A₃) while B, C and D appear once each. So

\(R_A = 3R_B = 3R_C = 3R_D\)

exactly as the statement claims. The A slots are also spread evenly through the frame, which is what a properly designed unequal-rate multiplexer does.

E — the same sequence A₁B₁A₂C₁A₃D₁ but claiming "all channels have the same bit rate." INCONSISTENT. A occupies three of the six slots, so it plainly cannot have the same rate as B, C and D. The sequence is fine; the rate claim contradicts it.

D — "A₁B₁C₁A₂D₁, channel A three times faster." INCONSISTENT. Here A gets only two slots against one each for B, C and D, i.e. a ratio of 2, not 3.

B — "A₁A₂B₁B₂C₁D₁C₂D₂, all channels at the same rate." INCONSISTENT with proper interleaving. Although the slot counts come out equal (two each), the samples are bunched in pairs rather than interleaved in a uniform rotation, so it does not represent the commutator sequence shown.

The consistent descriptions are A and C.

Key idea. To check any TDM frame, count each channel's slots per frame — that ratio is the ratio of bit rates. A faster channel is served by giving it extra, evenly spaced slots in every rotation.

Hence, the correct answer is A and C only.