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# Give answer! - Sequence and series - JEE Main-2

Let $a_1,a_2,a_3,.........$ be an A.P. with $a_6=2$ . Then the

common difference of this A.P., which maximises the product

$a_1a_4a_5$ , is :

• Option 1)

$\frac{3}{2}$

• Option 2)

• Option 3)

$\frac{6}{5}$

• Option 4)

$\frac{2}{3}$

Answers (1)
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Assuming the first term of A.P. is a and difference is d.

Then,

$a_{6}=a+5d=2$

Let $\Delta =a\cdot a_{4}\cdot a_5$

$\Delta =a.(a+3d).(a+4d)$

$\Delta =-2(5d^{3}-17d^{2}+16d-4)$

$\frac{d\Delta }{dd}=-2(15d^{2}-34d+16)$

=> $-2(5d-8)(3d-2)=0$

So, $\Delta$ will be maximum at $d=\frac{8}{5}$

So, option (2) is correct.

Option 1)

$\frac{3}{2}$

Option 2)

Option 3)

$\frac{6}{5}$

Option 4)

$\frac{2}{3}$

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