[(a_(0)+(A)/(a_(0))),a_(2)=(1)/(2)(a_(1)+(A)/(a_(1)))" and "a_(n+1)=(1)/(2)(a_(n)+(A)/(a_(n)))" for "n>=2],[[A>0],[-sqrt(A)],[a+sqrt(A)]=((a_(1)-sqrt(A))/(a_(1)+sqrt(A)))^(2^(n-1))]

Given a_(1)=(1)/(2)(a_(0)+(A)/(a_(0))),a_(2)=(1)/(2)(a_(1)+(A)/(a_(1))) and a_(n+1)=(1)/(2)(a_(n)+(A)/(a_(n))) for n>=2, where a>0,A>0. prove that (a_(n)-sqrt(A))/(a_(n)+sqrt(A))=((a_(1)-sqrt(A))/(a_(1)+sqrt(A)))2^(n-1)

If a_(1)=5 and a_(n)=1+sqrt(a_(n-1)), find a_(3) .

If a_(1), a_(2) …… a_(n) are in A.P then (sqrt(a_(1)) + sqrt(a_(n)))/( sqrt(a_(1)) + sqrt(a_(2))) + (sqrt(a_(1)) + sqrt(a_(n)))/(sqrt(a_(2)) + sqrt(a_(3))) + ….. + (sqrt( a_(1)) + sqrt(a_(n)))/( sqrt(a_(n - 1)) + sqrt(a_(n))) =

If a_(1),a_(2),a_(3),dots,a_(n+1) are in A.P.then (1)/(a_(1)a_(2))+(1)/(a_(2)a_(3))...+(1)/(a_(n)a_(n+1)) is

If a_(1),a_(2),a_(3),...,a_(2n+1) are in A.P.then (a_(2n+1)-a_(1))/(a_(2n+1)+a_(1))+(a_(2n)-a_(2))/(a_(2n)+a_(2))+...+(a_(n+2)-a_(n))/(a_(n+2)+a_(n)) is equal to (n(n+1))/(2)xx(a_(2)-a_(1))/(a_(n+1)) b.(n(n+1))/(2) c.(n+1)(a_(2)-a_(1)) d.none of these

If a_(1), a_(2), a_(3),….,a_(n) are in AP and a_(1) = 0 , then the value of ((a_(3))/(a_(2)) + (a_(4))/(a_(3)) +...+(a_(n))/a_(n-1))-a_(2) ((1)/(a_(2)) + (1)/(a_(3))+...+(1)/(a_(n-2))) is equal to a) (n-2) + (1)/((n-2)) b) (1)/((n-2)) c)(n - 2) d)(n - 1)

If a_(1),a_(2),a_(3),"........",a_(n) are in AP with a_(1)=0 , prove that (a_(3))/(a_(2))+(a_(4))/(a_(3))+"......"+(a_(n))/(a_(n-1))-a_(2)((1)/(a_(2))+(1)/(a_(3))"+........"+(1)/(a_(n-2))) = (a_(n-1))/(a_(2))+(a_(2))/(a_(n-1)) .

If a_(r ) gt 0, r in N and a_(1), a_(2) , a_(3) ,"……..",a_(2n) are in A.P. then : (a_(1) + a_(2n))/( sqrt( a_(1))+ sqrt(a_(2))) + ( a_(2) + a_(2n-1))/(sqrt(a_(2)) + sqrt(a_(3)))+"......"(a_(n)+a_(n+1))/(sqrt(a_(n))+sqrt(a_(n+1))) is equal to :

If a_(1) ge 0 for all t and a_(1), a_(2), a_(3),….,a_(n) are in A.P. then show that, (1)/(sqrt(a_(1))+sqrt(a_(2)))+(1)/(sqrt(a_(2))+sqrt(a_(3)))+ (1)/(sqrt(a_(n-1))+sqrt(a_(n))) = (n-1)/(sqrt(a_(1))+sqrt(a_(n)))

Let a_(1),a_(2),a_(3), . . .,a_(n) be an A.P. Statement -1 : (1)/(a_(1)a_(n))+(1)/(a_(2)a_(n-1))+(1)/(a_(3)a_(n-1))+ . . .. +(1)/(a_(n)a_(1)) =(2)/(a_(1)+a_(n))((1)/(a_(1))+(1)/(a_(2))+ . . .. +(1)/(a_(n))) Statement -2: a_(r)+a_(n-r+1)=a_(1)+a_(n)" for "1lerlen