If `A_1,A_2, A_3.........` are in `A.P.`, then `sum_(i=1)^(2n) (-1) ((A_1+A_(i+1))/(A_i-A_(i+1)))` is equal to `lambdan` then `|lambda|` is equal to

Let a_1,a_2 ,a_3 ...... be an A.P. Prove that sum_(n=1)^(2m)(-1)^(n-1)a_n^2=m/(2m-1)(a_1^2-a_(2m)^2) .

If a_1,a_2,a_3,... are in A.P. and a_i>0 for each i, then sum_(i=1)^n n/(a_(i+1)^(2/3)+a_(i+1)^(1/3)a_i^(1/3)+a_i^(2/3)) is equal to

If a_1,a_2,a_3,……………….. are in A.P., then the value of: {:|(a_1,a_2,1),(a_2,a_3,1),(a_3,a_4,1)| is equal to

If a_1,a_2,a_3,...,a_(n+1) are in A.P. , then 1/(a_1a_2)+1/(a_2a_3)....+1/(a_na_(n+1)) is

a_1,a_2,a_3 ,…., a_n from an A.P. Then the sum sum_(i=1)^10 (a_i a_(i+1)a_(i+2))/(a_i + a_(i+2)) where a_1=1 and a_2=2 is :

If a_1,a_2,a_3,... are in A.P. and a_i>0 for each i, then sum_(i=1)^n n/(a_(i+1)^(2/3)+a_(i+1)^(1/3)a_i^(1/3)+a_i^(2/3)) is equal to (a) (n)/(a_(n)^(2//3)+a_(n)^(1//3)+a_(1)^(2//3)) (b) (n(n+1))/(a_(n)^(2//3)+a_(n)^(1//3)+a_(1)^(2//3)) (c) (n(n-1))/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3)) (d) None of these

"If "a_1,a_2,a_3,.....,a_n" are in AP, prove that "a_(1)+a_(n)=a_(r)+a_(n-r+1)""

If a_1,a_2,……….,a_(n+1) are in A.P. prove that sum_(k=0)^n ^nC_k.a_(k+1)=2^(n-1)(a_1+a_(n+1))

If a_1, a_2, a_3,.....a_n are in H.P. and a_1 a_2+a_2 a_3+a_3 a_4+.......a_(n-1) a_n=ka_1 a_n , then k is equal to