If `a_na_(n-1)a-(n-2),.,a_3,a_2,a-1` be a digit having `a_1,a_2,a_3,…..a_n` unit, tens hundreds places respectively. Then `a_n a_(n-1) a_(n-2)….a_3a_2a_1= a_nxx1+a_2xx10^2+a_3xx10^3+…..+a_n10^(n-1)` On the basis of above information answer the following question If `alpha= 888.....8` (a nunmber of n digits), `beta=666.........6`(a number of n digits) and `gamma=444.......4` (a number 2n digits) then `gamma-alpha=` (A) `beta` (B) `beta^2` (C) -`beta` (D) `-beta^2`

If a_na_(n-1)a-(n-2),.,a_3,a_2,a-1 be a digit having a_1,a_2,a_3,…..a_n t unti, thens hundred places respectively. Then a_n a_(n-1) a_(n-2)….a_3a_2a_1= a_nxx1+a_2xx10^2+a_3xx10^3+…..+a_n10^(n-1) On the basis of above information answer the following question For a sequence {t_n},t_1=49, t_2= 4489, t_3=444889 in which every number is made by inserting 48 in the middle of the previous number. The for all nepsilonN, t_n is (A) square of an odd integer (B) divisible by 3 (C) divisible by 9 (D) none of these

Let a_1,a_2,a_3…………., a_n be positive numbers in G.P. For each n let A_n, G_n, H_n be respectively the arithmetic mean geometric mean and harmonic mean of a_1,a_2,……..,a_n On the basis of above information answer the following question: A_k,G_k,H_k are in (A) A.P. (B) G.P. (C) H.P. (D) none of these

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 :

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) =

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_n are an A.P. of non-zero terms, prove that 1/(a_1a_2)+1/(a_2a_3)++1/(a_(n-1)a_n)= (n-1)/(a_1a_n)

If a_n = (a_(n-1) xx a_(n-2))/(2), a_5 = -6 and a_6 = -18 , what is the value of a_3 ?

Write the first five terms in each of the following sequence: a_1=a_2=2,\ a_n=a_(n-1)-1,\ 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

If a_1,a_2,a_3,……a_n are in A.P. [1/(a_1a_n)+1/(a_2a_(n-1))+1/(a_3a_(n-2))+..+1/(a_na_1)]