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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`

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Given, `a_(1),a_(2),….,a_(n)` are in A.P., `AAa_(i)gt0`
`thereforea_(1)-a_(2)=a_(2)-a_(3)=….=A_(n-1)-a_(n)=-d` (a constant)
Now, `1/(sqrt(a_(1))+sqrt(a_(2)))+1/(sqrt(a_(2))+sqrt(a_(3)))+….+1/(sqrt(an-1)+sqrt(a_(n)))`
`=(sqrt(a_(1))-sqrt(a_(2)))/(a_(1)-a_(2))+(sqrt(a_(2))-sqrta_(3))/(a_(2)-a_(3))+...+(sqrt(a_(n-1))-sqrt(a_(n)))/(a_(n-1)-a_(n))`
(On rationalizing denominators)
`=(sqrt(a_(1))-sqrt(a_(2)))/(-d)+(sqrta_(2)-sqrta_(3))/(-d)+...+(sqrt(a_(n-1))-sqrt(a_(n)))/(-d)`
`=(sqrt(a_(n))-sqrt(a_(1)))/d`
`=(a_n-a_(1))/(d(sqrt(a_(n))+a_(a_(1))))`
`=((n-1)d)/(d(sqrt(a_(n))+sqrt(a_(1))))=(n-1)/(sqrt(a_(n))+sqrt(a_(1)))`
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