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Blazing goIITian

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24 May 2009 11:37:01 IST

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Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

The first term of an arithmetic progression is equal to unity . At what value of the common difference is

a₁a₃+a₂a₃ at a minimum . Give the full explanation.

Comments (5)

KAPIL CVS

Blazing goIITian

Joined: 7 May 2007

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24 May 2009 11:54:04 IST

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let the common difference = d be x.

first term =

so,

now,

f ' (x)

so,

now, f ' ' (x) = 4. it is positive. so, the value of x obtained is minima, and the value of f(x) should be minimum.

so, the common differene should be

Anirudh Kumar

Scorching goIITian

Joined: 25 Jan 2009

Posts: 246

24 May 2009 11:54:06 IST

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so a₁= 1

a₂= 1+d

a₃= 1+2d

now y= a₁a₃+a₃a₂ = 1+2d+ 1+d+2d+2d²= 2d²+5d+2

now , differentiating 'y' w.r.t to x and equating to 0(maxima and minima)

4d+5 = 0 thus d= -5/4

thusa₁a₃+a₃a₂ is minimum for d= -5/4

Anirudh Kumar

Scorching goIITian

Joined: 25 Jan 2009

Posts: 246

24 May 2009 11:55:10 IST

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hey this is wrong . i lost because of my slow typing.

KAPIL CVS

Blazing goIITian

Joined: 7 May 2007

Posts: 1724

24 May 2009 11:55:25 IST

Like 0 people liked this

let the common difference = d be x.

first term =

so,

now,

f ' (x)

so,

now, f ' ' (x) = 4. it is positive. so, the value of x obtained is minima, and the value of f(x) should be minimum.

so, the common differene should be

namita lohani

Hot goIITian

Joined: 14 Jul 2008

Posts: 178

24 May 2009 12:09:05 IST

Like 0 people liked this

given that,

a₁=1

now let us assume that common difference is d

hence ,

a₂=a₁+(2-1)d = 1+d

a₃=a₁+(3-1)d=1+2d

a₁a₃+a₂a₃=2+5d+2d²

we hav to minimize it....

diff. it with rspct to d ,we get

4d+5

now critical point is

f'(d) =0

d=-5/4

using 2nd derivative test

f"(d)=4>0

hence it refers to point of minima.therefore d=-5/4

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