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If sum(i=1)^(n)a(i)^(2)=lambda, AAa(i)ge...

If `sum_(i=1)^(n)a_(i)^(2)=lambda, AAa_(i)ge0` and if greatest and least values of `(sum_(i=1)^(n)a_(i))^(2)` are `lambda_(1)` and `lambda_(2)` respectively, then `(lambda_(1)-lambda_(2))` is

A

`nlambda`

B

`(n-1)lambda`

C

`(n+2)lambda`

D

`(n+1)lambda`

Text Solution

Verified by Experts

The correct Answer is:
B
`:.`AM of 2nd powes `ge` 2nd power of AM
`:.(a_(1)^(2)+a_(2)^(2)+a_(3)^(2)+"..."a_(n)^(2))/(n) ge((a_(1)+a_(2)+a_(3)+"..."a_(n))/(n))^(2)`
`implies (lambda)/(n)ge((sum_(i=1)^(n)a_(i))/(n))^(2) :.ge(sum_(i=1)^(n)a_(i))^(2)le n lambda " " ".......(i)"`
Also,`(a_(1)+a_(2)+a_(3)+"..."a_(n))^(2)=a_(1)^(2)+a_(2)^(2)+a_(3)^(2)+"..."a_(n)^(2)+2suma_(1)a_(2)=lambda+2suma_(1)a_(2)gelambda`
`:.(sum_(i=1)^(n)a_(i))^(2)le lambda" " ".......(ii)"`
From Eqs. (i) and (ii), we get
`lambdale(sum_(i=1)^(n)a_(i))^(2)le n lambda`
`:.lambda_(1)=nlambda" and "lambda_(2)=lambda`
Then, `lambda_(1)-lambda_(2)=(n-1)lambda`
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