If `a_1,a_2,……….,a_n` are in H.P. then expression a1a2+a2a3+......+a_n-1an

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

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

If a_1, a_2, ,a_n are in H.P., then (a_1)/(a_2+a_3++a_n),(a_2)/(a_1+a_3++a_n), ,(a_n)/(a_1+a_2++a_(n-1)) are in a. A.P b. G.P. c. H.P. d. none of these

Let the sequence a_1,a_2,a_3, ,a_n from an A.P. Then the value of a1 2-a2 2+a3 2-+a2n-1 2-a2n2 is (2n)/(n-1)(a2n2-a1 2) (b) n/(2n-1)(a1 2-a2n2) n/(n+1)(a1 2-a2n2) (d) n/(n-1)(a1 2+a2n2)

Let p_1,p_2,...,p_n and x be distinct real number such that (sum_(r=1)^(n-1)p_r^2)x^2+2(sum_(r=1)^(n-1)p_r p_(r+1))x+sum_(r=2)^n p_r^2 lt=0 then p_1,p_2,...,p_n are in G.P. and when a_1^2+a_2^2+a_3^2+...+a_n^2=0,a_1=a_2=a_3=...=a_n=0 Statement 2 : If p_2/p_1=p_3/p_2=....=p_n/p_(n-1), then p_1,p_2,...,p_n are in G.P.

If a_1,a_2,a_3,…………..a_n are in A.P. whose common difference is d, show tht sum_2^ntan^-1 d/(1+a_(n-1)a_n)= tan^-1 ((a_n-a_1)/(1+a_na_1))

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

The distributive law from algebra states that for all real numbers c,a1 and a2,we have c(a_1+a_2)=c a_1+c a_2 Use this law and mathematical induction to prove that,for all natural numbers, ngeq2 ,if c,a_1,a_2,...., an are any real numbers,then c(a_1+a_2+.....+a_n)=c a_1+c a_2+....+c a_n

cIf a_1,a_2,a_3,..,a_n in R then (x-a_1)^2+(x-a_2)^2+....+(x-a_n)^2 assumes its least value at x=

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