Let a_1,a_2,a_3... be an A.P. with a_6=2. Then, the common difference of this A.P., which maximise the product a_1· a_4· a_5,is :

Let, the first term of $A.P.$A.P. be $a$a and common difference $d,$d, then we know that the ${n}^{th}$nthterm of the $A.P.$A.P. is $a+(n-1)d.$a+(n−1)d.

Then ${a}_{6}=a+5d=2$a6​=a+5d=2

$$a=2-5d...(i)$$a=2−5d...(i)

Let $\Delta ={a}_{1}\cdot {a}_{4}\cdot {a}_{5}$Δ=a1​⋅a4​⋅a5​

$$\Rightarrow \Delta =a\cdot (a+3d)\cdot (a+4d)$$⇒Δ=a⋅(a+3d)⋅(a+4d)

Put the value of $a$a from equation $(i),$(i),

$$\Rightarrow \Delta =(2-5d)\cdot (2-5d+3d)\cdot (2-5d+4d)$$⇒Δ=(2−5d)⋅(2−5d+3d)⋅(2−5d+4d)

$$\Rightarrow \Delta =(2-5d)\cdot (2-2d)\cdot (2-d)$$⇒Δ=(2−5d)⋅(2−2d)⋅(2−d)

\Rightarrow \Delta =-2(5d3-17d2+16d-4)

For finding the maximum or minimum value of $\Delta$Δ

$$\Rightarrow \frac{d\Delta }{dd}=-2(15{d}^{2}-34d+16)$$⇒dddΔ​=−2(15d2−34d+16)

$$\Rightarrow \frac{d\Delta }{dd}=-2(5d-8)(3d-2)$$⇒dddΔ​=−2(5d−8)(3d−2)

Now, $\frac{d\Delta }{dd}=0$dddΔ​=0

$$\Rightarrow -2(5d-8)(3d-2)=0$$⇒−2(5d−8)(3d−2)=0

$\Rightarrow d=\frac{8}{5}$⇒d=58​ or $d=\frac{2}{3}$d=32​

And, $\frac{{d}^{2}\Delta }{d{d}^{2}}=-2(30d-34)$dd2d2Δ​=−2(30d−34)

At $d=\frac{8}{5},\frac{{d}^{2}\Delta }{d{d}^{2}}=-2(30\times \frac{8}{5}-34)=-28<0$d=58​,dd2d2Δ​=−2(30×58​−34)=−28<0 and at $d=\frac{2}{3},\frac{{d}^{2}\Delta }{d{d}^{2}}=-2(30\times \frac{2}{3}-34)=28>0$d=32​,dd2d2Δ​=−2(30×32​−34)=28>0

We know that if at any point the first derivative of a function is zero and the second derivative is negative, then it is the point of maxima of the function.

Hence, it is clear that $\Delta$Δ is maximum when $d=\frac{8}{5}.$d=58​.