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)

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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