If the roots of the equation $x^{3}+3px^{2}+3qx+r=0$ are in A.P, then the condition is:
- $2p^{3}=3pq+r$
- $2p^{3}=3pq$
- $2p^{3}+r=3pq$
- $2p^{3}-r=-3pq$
Since the roots are in arithmetic progression, they can be written as a - d, a, and a + d. Their sum is 3a = -3p, which gives a = -p. Substituting this root back into the original cubic equation yields (-p)^3 + 3p(-p)^2 + 3q(-p) + r = 0, which simplifies to 2p^3 + r = 3pq.
Let the roots of the cubic equation be a-d, a, and a+d since they are in arithmetic progression. The sum of the roots gives (a-d) + a + (a+d) = -3p, which simplifies to 3a = -3p and a = -p. The product of the roots is (a-d)a(a+d) = a(a^2 - d^2) = -r, so substituting a = -p yields p(p^2 - d^2) = r. The sum of the products of the roots taken two at a time is a(a-d) + a(a+d) + (a-d)(a+d) = 3q, which simplifies to 3a^2 - d^2 = 3q; substituting a = -p gives 3p^2 - d^2 = 3q, meaning d^2 = 3p^2 - 3q. Substituting this d^2 back into the product equation gives p(p^2 - (3p^2 - 3q)) = r, which is p(-2p^2 + 3q) = r and rearranges to 2p^3 + r = 3pq.