If `a_1, a_2, a_3...a_(2n-1)` are in H.P., then `sum_(k=1)^(2n)(-1)^k(a_k+a_(k+1))/(a_k-a_(k+1))` is equal to

If a_(1),a_(2),a_(3),…a_(n+1) are in arithmetic progression, then sum_(k=0)^(n) .^(n)C_(k.a_(k+1) is equal to (a) 2^(n)(a_(1)+a_(n+1)) (b) 2^(n-1)(a_(1)+a_(n+1)) (c) 2^(n+1)(a_(1)+a_(n+1)) (d) (a_(1)+a_(n+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+1) are in arithmetic progression, then sum_(k=0)^(n) .^(n)C_(k.a_(k+1) is equal to

If a_(a), a _(2), a _(3),…., a_(n) are in H.P. and f (k)=sum _(r =1) ^(n) a_(r)-a_(k) then (a_(1))/(f(1)), (a_(2))/(f (2)), (a_(3))/(f (n)) are in :

If a_(a), a _(2), a _(3),…., a_(n) are in H.P. and f (k)=sum _(r =1) ^(n) a_(r)-a_(k) then (a_(1))/(f(1)), (a_(2))/(f (2)), (a_(3))/(f (n)) are in :

If a_(a), a _(2), a _(3),…., a_(n) are in H.P. and f (k)=sum _(r =1) ^(n) a_(r)-a_(k) then (a_(1))/(f(1)), (a_(2))/(f (2)), (a_(3))/(f (n)) are in :

If a_(1),a_(2),a_(3),......,a_(n) are in A.P. Where a_(k)>0AA K and sum_(K=2)^(n)(1)/(sqrt(a_(K-1))+sqrt(a_(K)))=(L)/(sqrt(a_(1))+sqrt(a_(n))) then L=

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 a. (n(n+1))/2xx(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 H.P. and f(k) = sum_(r=1)^na_r-a_k then (f(1))/a_1, (f(2))/a_3 ….f(n)/a_n are (A) A.P. (B) G.P. (C) H.P. (D) none of these

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