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Question

Let a1,a2,a3,... be an A.P. with a6=2.Then the common difference of this A.P., which maximises the product a1a4a5, is :

A
23
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B
85
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C
65
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D
32
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Solution

The correct option is B 85
First term of the A.P.=a
Common difference =d
a+5d=2
a1a4a5=a(a+3d)(a+4d)=(25d)(22d)(2d)=2[5d3+17d216d+4]

Now, assuming
f(d)=2[5d3+17d216d+4]f(d)=2[15d2+34d16]f(d)=0(3d2)(5d8)=0d=23,85
Now,
f′′(d)=2[30d+34]f′′(d)=60(d1715)
At d=23, we get
f′′(d)>0
At d=85, we get
f′′(d)<0

Hence, the required common difference is 85

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