Match the $APs$ given in column $A$ with suitable common differences given in column $B$.

Column $A$	Column $B$
$(A_{1}) \quad 2, -2, -6, -10, \ldots$	$(B_{1}) \quad \frac{2}{3}$
$(A_{2}) \quad a = -18, n = 10, a_{n} = 0$	$(B_{2}) \quad -5$
$(A_{3}) \quad a = 0, a_{10} = 6$	$(B_{3}) \quad 4$
$(A_{4}) \quad a_{2} = 13, a_{4} = 3$	$(B_{4}) \quad -4$
	$(B_{5}) \quad 2$
	$(B_{6}) \quad \frac{1}{2}$
	$(B_{7}) \quad 5$

(A-B) $A_{1}: 2, -2, -6, -10, \ldots$
Here,the common difference $d = a_{2} - a_{1} = -2 - 2 = -4$. Thus,$(A_{1}) \rightarrow (B_{4})$.
$A_{2}: a = -18, n = 10, a_{n} = 0$. Using $a_{n} = a + (n - 1)d$:
$0 = -18 + (10 - 1)d \Rightarrow 18 = 9d \Rightarrow d = 2$. Thus,$(A_{2}) \rightarrow (B_{5})$.
$A_{3}: a = 0, a_{10} = 6$. Using $a_{n} = a + (n - 1)d$:
$a_{10} = a + 9d \Rightarrow 6 = 0 + 9d \Rightarrow d = \frac{6}{9} = \frac{2}{3}$. Thus,$(A_{3}) \rightarrow (B_{1})$.
$A_{4}: a_{2} = 13, a_{4} = 3$. Using $a_{n} = a + (n - 1)d$:
$a + d = 13$ $(i)$
$a + 3d = 3$ $(ii)$
Subtracting $(i)$ from $(ii)$: $(a + 3d) - (a + d) = 3 - 13 \Rightarrow 2d = -10 \Rightarrow d = -5$. Thus,$(A_{4}) \rightarrow (B_{2})$.
Final matching: $(A_{1}) \rightarrow (B_{4}), (A_{2}) \rightarrow (B_{5}), (A_{3}) \rightarrow (B_{1}), (A_{4}) \rightarrow (B_{2})$.