The correct option is β⁺, α, β⁻.

Let's analyze the given radioactive decay sequence step-by-step to identify the emitted particles. Let the initial nucleus be represented as:
$X_Z^A$ where A is the mass number and Z is the atomic number.

Step 1: First Decay
The first transition is from element X with atomic number Z to element B with atomic number Z - 1:
$X_Z^A\to B_{Z-1}^{A_1}+\text{particle}$
Here, the atomic number decreases by 1 (from Z to Z - 1). This decrease in atomic number by 1 corresponds to positron emission, also known as beta-plus decay (β⁺), where a proton turns into a neutron, emitting a positron ($e_{+1}^0$) and a neutrino. Therefore, the first decay particle is β⁺.

Step 2: Second Decay
The second transition is from element B with atomic number Z - 1 to another state or isotope of element B (or another element) with atomic number Z - 3:
$B_{Z-1}^{A_1}\to C_{Z-3}^{A_2}+\text{particle}$
Here, the atomic number decreases by 2 (from Z - 1 to Z - 3). A decrease in the atomic number by 2 is characteristic of alpha decay (α), in which an alpha particle (${\mathrm{He}}_2^4$) is emitted. Therefore, the second decay particle is α.

Step 3: Third Decay
The third transition is to element D with atomic number Z - 2:
$C_{Z-3}^{A_2}\to D_{Z-2}^{A_3}+\text{particle}$
Here, the atomic number increases by 1 (from Z - 3 to Z - 2). An increase in the atomic number by 1 is characteristic of beta-minus decay (β⁻), where a neutron decays into a proton, emitting an electron ($e_{-1}^0$) and an antineutrino. Therefore, the third decay particle is β⁻.

Combining the results of the three steps, the sequence of emitted decay particles is β⁺, α, β⁻.