In an A.P., the sixth term a₆ = 2. If the product a₁a₄a₅ is the greatest, then the common difference of the A.P. is equal to

The correct answer is 8/5.

We are given an Arithmetic Progression (A.P.) where the sixth term $a_6=2$. We need to find the common difference $d$ such that the product $a_1{a}_4{a}_5$ is maximized.

Step 1: Express all terms using the first term $a$ and common difference $d$.

The general term of an A.P. is $a_n=a+(n-1)d$.

From the given condition $a_6=2$:

$a+5d=2$    ⇒    $a=2-5d$

Now we express each required term:

$a_1=a=2-5d$
$a_4=a+3d=2-5d+3d=2-2d$
$a_5=a+4d=2-5d+4d=2-d$

Step 2: Set up the product function P(d).

$P=a_1·a_4·a_5=(2-5d)(2-2d)(2-d)$

First, expand the first two factors:

$(2-5d)(2-2d)=4-4d-10d+10d^2=4-14d+10d^2$

Then multiply by $(2-d)$:

$P=(4-14d+10d^2)(2-d)$

$P=8-4d-28d+14d^2+20d^2-10d^3$

$P=8-32d+34d^2-10d^3$

Step 3: Differentiate P with respect to d and set equal to zero.

$\frac{\mathrm{dP}}{\mathrm{dd}}=-32+68d-30d^2=0$

Multiplying through by -1 and rearranging:

$30d^2-68d+32=0$

Dividing the entire equation by 2:

$15d^2-34d+16=0$

Step 4: Solve the quadratic using the quadratic formula.

$d=\frac{34\pm \sqrt{{34}^2-4·15·16}}{2·15}=\frac{34\pm \sqrt{1156-960}}{30}=\frac{34\pm \sqrt{196}}{30}=\frac{34\pm 14}{30}$

This gives two critical points:

$d=\frac{34+14}{30}=\frac{48}{30}=\frac{8}{5}$

$d=\frac{34-14}{30}=\frac{20}{30}=\frac{2}{3}$

Step 5: Apply the Second Derivative Test to determine which critical point gives a maximum.

$\frac{d^{2}P}{d{d}^2}=68-60d$

For $d=\frac{8}{5}$:   $68-60·\frac{8}{5}=68-96=-28<0$   → Maximum ✓

For $d=\frac{2}{3}$:   $68-60·\frac{2}{3}=68-40=28>0$   → Minimum ✗

Since the second derivative is negative at $d=\frac{8}{5}$, this is a point of maximum. Therefore, the product $a_1{a}_4{a}_5$ is greatest when the common difference $d=\frac{8}{5}$.