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. $\frac{8}{5}$
• B. $\frac{5}{4}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

Answer: (C)

$\frac{2}{3}$

Given that sixth term of an A.P is 2.

$\Rightarrow a_1+5d=2$

Consider, $p=a_1{a}_4{a}_5$

$\Rightarrow p=a_1(a_1+3d)(a_1+4d)$

Now substitute $a_1=2-5d$ in the above equation, we get;

$p=(2-5d)(2-2d)(2-d)$

Now, solving the parentheses, we get;

$p=2[4-16d+17d^2-5d^3]$

Let, $S=-5d^3+17d^2-16d+4$

Now, taking the derivative of S w.r.t $d$, we get;

$S\prime =-15d^2+34d-16$

Notice that, for $S\prime =0$, we get;

$d=\frac{2}{3},\frac{8}{5}$

Now, taking the derivative of $S\prime$ w.r.t $d$, we get;

$S\prime \prime =-30d+34$

At $d=\frac{2}{3}$, we get;

$S\prime \prime =-20+34$

So, $S\prime \prime =14$ which is positive.

Therefore, $d=\frac{2}{3}$ gives minimum value.