Let `a_1, a_2, a_3, ...` be a G.P such that `log_(10) (a_m)=1/n` and `log_(10) (a_n)=1/n` then

Let a_1, a_2, a_3, ...a_(n) be an AP. then: 1 / (a_1 a_n) + 1 / (a_2 a_(n-1)) + 1 /(a_3a_(n-2))+......+ 1 /(a_(n) a_1) =

Let a_1,a_2,a_3 ......, a_n are in A.P. such that a_n = 100 , a_40-a_39=3/5 then 15^(th) term of A.P. from end is

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 G.P. then the value of determinant |(log(a_n), log(a_(n+1)), log(a_(n+2))), (log(a_(n+3)), log(a_(n+4)), log(a_(n+5))), (log(a_(n+6)), log(a_(n+7)), log(a_(n+8)))| equals (A) 0 (B) 1 (C) 2 (D) 3

Let a_1,a_2,a_3…… ,a_n be in G.P such that 3a_1+7a_2 +3a_3-4a_5=0 Then find common ratio of G.P.

Let a_1,a_2,a_3,.... are in GP. If a_n>a_m when n>m and a_1+a_n=66 while a_2*a_(n-1)=128 and sum_(i=1)^n a_i=126 , find the value of n .

Let a_1,a_2,a_3…… ,a_n be in G.P such that 3a_1+7a_2 +3a_3-4a_5=0 Then common ratio of G.P can be

Let a_1,a_2,a_3…… ,a_n be in G.P such that 3a_1+7a_2 +3a_3-4a_5=0 Then common ratio of G.P can be

Let a_1,a_2,a_3…… ,a_n be in G.P such that 3a_1+7a_2 +3a_3-4a_5=0 Then common ratio of G.P can be