In an A.P., the sixth term $$a_6 = 2$$. If the $$a_1 a_4 a_5$$ is the greatest, then the common difference of the A.P., is equal to

Express terms in terms of $$a_6$$ and $$d$$:

• $$a_5 = 2 - d$$

• $$a_4 = 2 - 2d$$

• $$a_1 = 2 - 5d$$

Let $$P(d) = (2-5d)(2-2d)(2-d)$$.

$$P(d) = (2-5d)(4 - 6d + 2d^2) = -10d^3 + 34d^2 - 32d + 8$$.

To maximize, find $$P'(d) = 0$$:

$$P'(d) = -30d^2 + 68d - 32 = 0 \implies 15d^2 - 34d + 16 = 0$$.

Using the quadratic formula:

$$d = \frac{34 \pm \sqrt{1156 - 960}}{30} = \frac{34 \pm 14}{30}$$

$$d_1 = \frac{48}{30} = \frac{8}{5}$$ and $$d_2 = \frac{20}{30} = \frac{2}{3}$$.

Check $$P''(d) = -60d + 68$$.

For $$d = 8/5$$, $$P''(8/5) = -96 + 68 < 0$$ (Maximum).

Answer: B ($$8/5$$)