Let `a_(1), a_(2),a_(3), cdots , a_(4001)` are in A.P. such that `(1)/(a_(1)a_(2))+(1)/(a_(2)a_(3))+ cdots + (1)/(a_(4000) a_(4001))= 10` and `a_(2)+ a_(4000) = 50` then find the value of `|a_(1)-a_(4001)|`.

If a_(n+1)= (1)/(1-a_(n)) for n gt=1 and a_(3) = a_(1) then find the value of (a_(2001))^(2001) .

If a_(1), a_(2),a_(3), cdots , 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))+cdots+ (a_(n+_2)-a_(n))/(a_(n+2)+a_(n)) is equal to

If a_(1), a_(2), a_(3), a_(4) are in AP, then (1)/(sqrt(a_(1)) + sqrt(a_(2))) + (1)/(sqrt(a_(2)) + sqrt(a_(3))) + (1)/(sqrt(a_(3)) + sqrt(a_(4))) is equal to

If a_1 = 4 and a_(n+1) = a_(n) + 4n for n ge 1 , then the value of a_(100) is

If a_(1) , a_(2), a_(3) , cdots ,a_(n) are in A.P. with a_(1) =3, a_(n) =39 and a_(1) +a_(2) + cdots +a_(n) =210 then the value of n is equal to

Let a_(1) ,a_(2) ,a_(3) ,cdots ,a_(n) are in A.P such that a_(n) =100 , a_(10)- a_(39 ) = (3)/(5) then find the 15^(th) term of A.P. from end.

Let a_(1) ,a_(2),cdots , a_(n) be an A.P. with common difference pi//6 and assume sec a_(1) sec a_(2)+sec a_(2) sec a_(3) + cdots + sec a_(n-1) sec a_(n)=k ( tan a_(n) - "tan" a_(1) ) Then find the value of k.

If a_(1), a_(2), a_(3),…….,a_(20) are in AP and a_(1) + a_(20) = 45 , then a_(1) + a_(2) + a_(3)+……+a_(20) is equal to

If a_(1), a_(2), …, a_(50) are in GP, then " "(a_(1)-a_(3)+a_(5)-…+a_(49))/(a_(2)-a_(4)+a_(6)-…+a_(50)) is equal to :

Let a_(1),a_(2),a_(3),a_(4) and a_(5) be such that a_(1), a_(2) and a_(3) are in A.P. a_(2) ,a_(3) and a_(4) are in G.P and a_(3) ,a_(4) and a_(5) are in H.P . Then a_(1) ,a_(3) and a_(5) are in