If `a_1,a_2,...a_n` are in H.P then the expression `a_1a_2+a_2a_3+....+a_ (n-1)a_n` is equal to

Let the sequence a_1 , a_2 , a_3 ......... a_n form an A.P. then a_1^2 - a_2^2 + a_3^2 - a_4^2 +.....+ a_(2n-1)^2 - a_(2n)^2 is equal to:- (1) n/(2n-1)(a_1^2-a_(2n)^2) (2) (2n)/(n-1)(a_(2n)^2-a_1^2) (3) n/(n+1)(a_1^2+a_(2n)^2) (4)none of these

Let the sequence a_1 , a_2 , a_3 ......... a_n form an A.P. then a_1^2 - a_2^2 + a_3^2 - a_4^2 +.....+ a_(2n-1)^2 - a_(2n)^2 is equal to:- (1) n/(2n-1)(a_1^2-a_(2n)^2) (2) (2n)/(n-1)(a_(2n)^2-a_1^2) (3) n/(n+1)(a_1^2+a_(2n)^2) (4)none of these

If a_(n) = n(n!) , then what is a_1 +a_2 +a_3 +......+ a_(10) equal to ?

If a_1,a_2,a_3,.....,a_n are in AP, prove that 1/(a_1a_2)+1/(a_2a_3)+1/(a_3a_4)+...+1/(a_(n-1)a_n)=(n-1)/(a_1a_n) .

"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 the nonzero numbers a_1,a_2,a_3,....,a_n are in AP, prove that 1/(a_1a_2a_3)+1/(a_2a_3a_4)+...+1/(a_(n-2)a_(n-1)a_n)=1/(2(a_2-a_1))(1/(a_1a_2)-1/(a_(n-1)a_n)) .

We know that, if a_1, a_2, ..., a_n are in H.P. then 1/a_1,1/a_2,.....,1/a_n are in A.P. and vice versa. If a_1, a_2, ..., a_n are in A.P. with common difference d, then for any b (>0), the numbers b^(a_1),b^(a_2),b^9a_3),........,b^(a_n) are in G.P. with common ratio b^d. If a_1, a_2, ..., a_n are positive and in G.P. with common ration, then for any base b (b> 0), log_b a_1 , log_b a_2,...., log_b a_n are in A.P. with common difference logor.If x, y, z are respectively the pth, qth and the rth terms of an A.P., as well as of a G.P., then x^(z-x),z^(x-y) is equal to

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