If the sixth term of an increasing A.P. is equal to 2, then the value of the common difference of the A.P. such that the product $a_{1} a_{4} a_{5}$ is greatest, is (the $i^{t h}$ term is denoted by $a_{i}$ )

$\frac{8}{5}$

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$\frac{2}{3}$

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Solution

The correct option is A
$\frac{8}{5}$

$a_{1} + 5 x = 2$

$\Rightarrow a_{1} = 2 - 5 x$

Now, $a_{1} . a_{4} . a_{5} = (2 - 5 x) (2 - 5 x + 3 x) (2 - 5 x + 4 x)$

$= (2 - 5 x) (2 - 2 x) (2 - x)$

$= 8 - 12 x + 4 x^{2} - 20 x + 30 x^{2} - 10 x^{3}$

$p = 8 - 32 x + 34 x^{2} - 10 x^{3}$

$\Rightarrow \frac{d p}{d x} = - 32 + 68 x - 30 x^{2} = 0$

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

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

for $x = \frac{8}{5}, \frac{d^{2} p}{d x^{2}} < 0$

$∴$ Common difference = $\frac{8}{5}$ .

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