If `a_1,a_2,a_3, a_n` are in arithmetic progression with common difference `d ,` then evaluate the following expression: `tan{tan^(-1)(d/(1+a_1a_2))+tan^(-1)(d/(1+a^2a_3))+tan^(-1)(d/(1+a_3a_4))++tan^(-1)(d/(1+a_(n-1)a_n))}`

If a_(1),a_(2),a_(3),a_(n) is an A.P.with common difference d, then prove that tan[tan^(-1)((d)/(1+a_(1)a_(2)))+tan^(-1)((d)/(1+a_(2)a_(3)))+tan^(-1)((d)/(1+a_(n-1)a_(n)))]=((n-1)d)/(1+a_(1)a_(n))

If a_(1), a_(2), a_(3),...., a_(n) is an A.P. with common difference d, then prove that tan[tan^(-1) ((d)/(1 + a_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((d)/(1 + a_( - 1)a_(n)))] = ((n -1)d)/(1 + a_(1) a_(n))

If a_(1), a_(2), a_(3) are in arithmetic progression and d is the common diference, then tan^(-1)((d)/(1+a_(1)a_(2)))+tan^(-1)((d)/(1+a_(2)a_(3)))=

tan^-1 (1/(1+2))+tan^-1(1/(1+(2)(3)))tan^-1(1/(1+(3)(4)))+...tan^-1(1/(1+n(n+1)))=tan^-1 theta

If a_(1),a_(2),a_(3),….a_(n) is a.p with common difference d then tan{tan^(-1)((d)/(1+a_(1)a_(2)))+tan^(-1)((d)/(1+a_(2)a_(3))) +..+ tan^(-1)((d)/(1+a_(n-1)a_(n)))} is equal to

tan^(-1)((n)/(n+1))-tan^(-1)(2n+1)=(3 pi)/(4)

tan^(-1)((3)/(n))+tan^(-1)((4)/(n))=(pi)/(2)

If a_1,a_2,a_3,…….a_n are in Arithmetic Progression, whose common difference is an integer such that a_1=1,a_n=300 and n in[15,50] then (S_(n-4),a_(n-4)) is