1. **State the problem:** We have an arithmetic progression (A.P.) with first term $a_1 = 6$. 2. The 10th term is equal to three times the second term. We need to find the common difference $d$. 3. Recall the formula for the $n$th term of an A.P.: $$a_n = a_1 + (n-1)d$$ 4. Write the expression for the 10th term: $$a_{10} = 6 + 9d$$ 5. Write the expression for the second term: $$a_2 = 6 + d$$ 6. Set up the equation from the problem statement: $$a_{10} = 3a_2$$ So, $$6 + 9d = 3(6 + d)$$ 7. Expand the right side: $$6 + 9d = 18 + 3d$$ 8. Simplify and solve for $d$: $$9d - 3d = 18 - 6$$ $$6d = 12$$ $$d = 2$$ 9. **Answer:** The common difference is $2$. Hence, the correct option is (d) 2.