If b(1),b(2),b(3),"….."b(n) are positive...

If `b_(1),b_(2),b_(3),"….."b_(n)` are positive then the least value of `(b_(1) + b_(2) +b _(3) + "….." + b_(n)) ((1)/(b_(1)) + (1)/(b_(2)) + "….." +(1)/(b_(n)))` is

`b_(1)b_(2)"…."b_(n)`

`n^(2) + 1`

`n(n+1)`

`n^(2)`

Text Solution

Verified by Experts

The correct Answer is:

(i) Take `b_(1)=b_(2)=b_(03)"……."b_(n) = k` and find the value .
(ii) AM `(a_(1),a_(2),"……"a_(n)) ge HM (a_(1),a_(2),"…….",a_(n))`
(iii) `(a_(1)+a_(2)+"….."+a_(n)).(n) ge (n)/((1)/(a_(1))+ (1)/(a_(2)) + (1)/(a_(2)))`.

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