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`

If `a_1,a_2,a_3, ,a_n`are in A.P., where `a_i >0`for all `i`, show tha

1.	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`
Answer» Let d be the common difference of the given AP. Then , ` ( a_(2) -a_(1)) = (a_(3) -a_(2)) = …..= (a_(n) -a_(n-1))= d ` Now , LHS ` 1/ ( (sqrta_(2) + sqrta_(1))) + 1 / (( sqrta_(3) + sqrta_(2)))+…..+ 1/((sqrta_(n) + sqrta_(n-1)))` ` ((sqrta_(2) - sqrta_(1)))/((sqrta_(2) +sqrta_(1))sqrta_(2) -sqrta_(1))) + ((sqrta_(3)-sqrta_(2)))/((sqrta_(3) + sqrta_(2)) (sqrta_(3) -sqrta_(3) -sqrta_(2))) +......+ ((sqrta_(n) -sqrta_(n-1)))/((sqrta_(n)+sqrt(a_(n-1))) (sqrta_(n) -sqrt(a_(n-1)))` `((sqrta_(2)-sqrta_(1)))/((a_(2)-a_(1)))+((sqrta_(3)-sqrta_(2)))/((a_(3)-a_(2)))+.......+((sqrta_(n)-sqrta_(n-1)))/((a_(n) -a_(n-1)))` ` ((sqrta_(2) -sqrta_(1))/d + ((sqrta_(3))-sqrta_(2))/d+....+ ((sqrta_(n) -sqrta_(n-1)))/d` `[ therefore (a_(2) -a_(1)) = (a_(3) -a_(2)) = (a_(4) -a_(3)) = .......= (a_(n) -a_(n-1)) =d]` ` 1/d .(sqrta_(n) -sqrt(a_(1)) + (sqrta_(3) -sqrt(a_(2)) + (sqrta_(4)-sqrta_(3)) +.....+ (sqrta_(n) -sqrta_(n-1))} ` =` 1/d .(sqrta_(n) -sqrt(a_(1))= 1/d .{(sqrta_(n) -sqrt(a_(1))xx ((sqrta_(n) +sqrta_(1)))/(sqrta_(n) + sqrt(a_(1)))}` ` 1/d . {((a_(n)-a_(1))/((sqrta_(1) +sqrta_(n))] = 1/d. {(a_(1) + (n+1)_ d-a_(1))/(sqrta_(1) + sqrta_(n))}` ` [ a_(n) =a_(1) + ( n-1) d]` `1/d .{ ((n-1) d)/( sqrta_(1) +sqrta_(n)))} = ((n-1))/((sqrta_(1)+sqrta_(n))) `= RHS Thus, LHS=RHS and hence the result follows.