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If a1,a2,a3, ,an are in A.P., where ai >...

If `a_1,a_2,a_3, ,a_n` are in A.P., where `a_i >0` for all `i` , show that `1/(sqrt(a_1)+sqrt(a_2))+1/(sqrt(a_1)+sqrt(a_3))++1/(sqrt(a_(n-1))+sqrt(a_n))=(n-1)/(sqrt(a_1)+sqrt(a_n))dot`

A

` (n^(2)(n+1))/2`

B

` (n-1)/(sqrta_(1) +sqrta_(n))`

C

` (n(n-1))/2`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
b
Since , `a_(1),a_(2),….,a_(n)` are I Ap.
Then, `a_(1)-a_(1)=a_(3)= …=a_(n)-a_(n-1) =d`
where , d is common difference .
Now, `(1)/(sqrt(a_(2))+sqrt(a_(1)))+(1)/(sqrt(a_(3))+sqrt(a_(2)))+...+(1)/(sqrt(a_(n))+sqrt(a_(m-1)))`
`=(sqrt(a_(2))-sqrt(a_(1)))/(d)+(sqrt(a_(3))-sqrt(a_(2)))/(d)+...+(sqrt(a_(n))-sqrt(a_(n-1)))/(d)`
`=(1)/(d)(sqrt(a_(n))-sqrt(a_(1)))xx(sqrt(a_(n))+sqrt(a_(1)))/(sqrt(a_(n))+sqrt(a_(1)))`
`=(1)/(d)((a_(n)-a_(1))/(sqrt(a_(n))+sqrt(a_(1))))=(n-1)/(sqrt(a_(n))+sqrt(a_(1)))`
`" "[because a_(n)=a_(1)+(n-1)d]`
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