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. $x = 2 / 3$
• D. None of these

Answer 1

Answer: (C)

$x = 2 / 3$

Answer 2

Let $a$ be the first term and $x$ be the common difference of the A.P. Then $a + 5 x = 2$ $\Rightarrow$ $a = 2 - 5 x$

Let$P = a _ { 1 } a _ { 4 } a _ { 5 } = a ( a + 3 x ) ( a + 4 x )$

$= ( 2 - 5 x ) ( 2 - 2 x ) ( 2 - x ) = 2 \left( - 5 x ^ { 3 } + 17 x ^ { 2 } - 16 x + 4 \right)$

Now $\frac { d P } { d x } = 0$ $\Rightarrow$ $x = \frac { 8 } { 5 } , \frac { 2 } { 3 }$ .

Clearly, $\frac { d ^ { 2 } P } { d x ^ { 2 } } > 0$ for $x = \frac { 2 } { 3 }$

Hence $P$ is least for $x = \frac { 2 } { 3 }$.