Question

The sixth term of an A.P is equal to $2$, the value of the common difference of the A.P. which makes the product $a_1{a}_4{a}_5$ least is given by

• A. $x=\frac{8}{5}$
• B. $x=\frac{5}{4}$
• C. $x=\frac{2}{3}$
• D. none of these

Answer 1

The correct option is C

$x=\frac{2}{3}$

Explanation for the correct option:

Step 1: Find the required product of terms of A.P.

Let $a$be the first term and $d$ be the common difference of the AP.

Given 6th term$=2$

$$\begin{gathered} \Rightarrow a+5d=2 \\ \Rightarrow a=2-5d \end{gathered}$$

Let the product of $a_1{a}_4{a}_5$ is $P$

$$\begin{gathered} P=a(a+3d)(a+4d) \\ =(2-5d)(2-5d+3d)(2-5d+4d) \\ =(2-5d)(2-2d)(2-d) \\ =2(1-d)(2-d)(2-5d) \\ \therefore P=2(-5d^3+17d^2-16d+4) \end{gathered}$$

Step 2: Find the minimum value of the above product.

For minimum value of$P,P\text{'}\left(d\right)=0$

Differentiate $P$ w.r.t. $d$, we get

$P\text{'}\left(d\right)=2(-15d^2+34d-16)$

$$\begin{gathered} \therefore 2(-15d^2+34d-16)=0 \\ \Rightarrow (-15d^2+34d-16)=0 \\ \Rightarrow (3d-2)(5d-8)=0 \\ \Rightarrow d=\frac{2}{3}ord=\frac{8}{5} \end{gathered}$$

Again differentiate w.r.t. $d$, we get

$P\text{'}\text{'}\left(d\right)=2(-30d+34)$

Now at $d=\frac{2}{3}$

$P\text{'}\text{'}\left(d\right)=28$ [ positive ]

and at $d=\frac{8}{5}$

$P\text{'}\text{'}\left(d\right)=-28$ [ negative ]

We get minimum value when $P\text{'}\text{'}\left(d\right)>0$.

So $d=\frac{2}{3}$ gives the least value.

Hence, option $C$ is the correct answer.