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`

Explanation

Since `a_(1) ,a_(2) ,a_(3)…a_(n) ` are in AP
`implies a_(2) -a_(1) =a_(3)-a_(2) =.= a_(n) -a_(n-1)=d [ " common difference "]`
if `A_(2)-a_(1)=d ` then `(sqrt(a_(2)))^(2) -(sqrt(a_(1)))^(2)=d`
`implies (sqrt(a_(2))-sqrt(a_(1)))(sqrt(a_(2))+sqrt(a_(1))=d`
`(1) /(sqrt(a_(1))+sqrt(a_(3))) =(sqrt(a_(2))-sqrt(a_(1)))/(d)`
` "Similarly ", (1) /(sqrt(a_(2))+sqrt(a_(3)))=(sqrt(a_(3))-sqrt(a_(2) ))/(d)`
`{:(...,...,...),(...,...,...),(...,...,...):}`
` (1)/(sqrt(a_(n-1))+sqrt(a_(n)))=(sqrt(a_(n))sqrt(a_(n-1)))/(d)`
On adding these terms we get
`(1)/(sqrt(a_(1))+sqrt(a_(2)))+(1)/(sqrt(a_(2))+sqrt(a_(3)))+....+ (1)/(sqrt(a_(n-1))+sqrt(a_(n)))`
` (1)/(d) [sqrt(a_(2))-sqrt(a_(1))+sqrt(a_(3))-sqrt(a_(2))+...+sqrt(a_(n))-sqrt(a_(n-1))]`
` =(1)/(d) [sqrt( a_(n)=-sqrt(a_(1))]`
Again ` a_(n)a_(1)+(n-1)d`
`implies a_(n)-a_(1)=(n-1)d`
` implies (sqrt(a_(n)))^(2) -(sqrt(_(1)))^(2)=(n-1)d`
` implies (sqrt(a_(n)))-sqrt(a_(1))(sqrt(a_(n))+sqrt(a))=(n-1)dimplies sqrt(a_(n))-sqrt(a_(1))=((n-1)d)/(sqrt(a_(n))+sqrt(a_(1)))`
On puting this value in eq (i)we get
`(1)/(sqrt(a_(1))+sqrt(a_(2)))+(1)/(sqrt(a_(2))+sqrt(a_(3)))+...+ (1)/(sqrt(a_(n-1))+sqrt(a_(n)))`
` =((n-1)d)/(d(sqrta_(n)+sqrt(a_(1))))=(n-1)/(sqrt(a_(n))+sqrt(a_(1)))` hence proved .