If `a_1, a_2, a_n` are in A.P. with common difference `d!=0` , then the sum of the series `sind[seca_1seca_2+(sec)_2seca_3+....+s e ca_(n-1)(sec)_n]` is : a.`cos e ca_n-cos e ca` b. `cota_n-cota` c. `s e ca_n-s e ca` d. `t a na_n-t a na`

If a_(1), a_(2), …..,a_(n) are in A.P. with common difference d ne 0, then the sum of the series sin d[sec a_(1)sec a_(2) +..... sec a_(n-1) sec a_(n)] is

If a_1, a_2, a_3,...a_n are in A.P with common difference d !=0 then the value of sind(coseca_1 coseca_2 +cosec a_2 cosec a_3+...+cosec a_(n-1) cosec a_n) will be

let a_1,a_2,...a_n be in A.P. wth common difference pi/6. if seca_1seca_2+seca_2seca_3+....seca_(n-1)seca_n=k(tana_n-tana_1)Find the value of k

If a_1,a_2,a_3,...,a_n are in AP and a_i ne (2k-1)pi/2 for all i , find the sum seca_1*seca_2+seca_2*seca_3+seca_3*seca_4+...+seca_(n-1)*seca_n .

If a_1,a_2,a_3,...,a_n be in AP whose common difference is d then prove that sum_(i=1)^n a_ia_(i+1)=n{a_1^2+na_1d+(n^2-1)/3 d^2} .

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_n))

If a_1,a_2,a_3,.....,a_n are in AP where a_i ne kpi for all i , prove that cosec a_1* cosec a_2+ cosec a_2* cosec a_3+...+ cosec a_(n-1)* cosec a_n=(cota_1-cota_n)/(sin(a_2-a_1)) .

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