let `a_0,a_1,a_2.... `be in A.P. STATEMENT 1: `sina_2+sina_4+....sina_(2n)=(cos a_1-cos a_(2n+1))/(2 sin(a_2-a_1))` because STATEMENT `2:sum_(r=1)^ncosa_(2r)=(cosa_(2n+1)-cosa_1)/(2cos(a_2-a_1))`

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) .

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_(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))

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_1, a_2, a_3, be terms of an A.P. if (a_1+a_2+...+a_p)/(a_1+a_2+...+a_q)=(p^2)/(q^2), p!=q ,t h e n(a_6)/(a_(21)) equals

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

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

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 AP, prove that "a_(1)+a_(n)=a_(r)+a_(n-r+1)""