If `a_(1),a_(2),a_(3),……a_(87),a_(88),a_(89)` are the arithmetic means between `1` and `89`, then `sum_(r=1)^(89)log(tan(a_(r ))^(@))` is equal to

If a_(1),a_(2),a_(3)...a_(87),a_(88),a_(89) are the arithmetic means between 1 and 89 then sum^(89)log(tan(a_(r))^(@)) is equal to

If a_(1), a_(2), a_(3), a_(4), a_(5) and a_(6) are six arithmetic means between 3 and 31, then a_(6) - a_(5) and a_(1) + a_(6) are respectively =

If A_(1), A_(2), A_(3),....A_(51) are arithmetic means inserted between the number a and b, then find the value of ((b + A_(51))/(b - A_(51))) - ((A_(1) + a)/(A_(1) - a))

If A_(1), A_(2), A_(3),....A_(51) are arithmetic means inserted between the number a and b, then find the value of ((b + A_(51))/(b - A_(51))) - ((A_(1) + a)/(A_(1) - a))

If a_(1),a_(2),a_(3),..., are in arithmetic progression,then S=a_(1)^(2)-a_(2)^(2)+a_(3)^(2)-a_(4)^(2)+...-a_(2k)^(2) is equal

If n arithmetic means A_(1),A_(2),---,A_(n) are inserted between a and b then A_(2)=

Statement-1: If f:{a_(1),a_(2),a_(3),a_(4),a_(5)}to{a_(1),a_(2),a_(3),a_(4),a_(5)} , f is onto and f(x)nex for each x in{a_(1),a_(2),a_(3),a_(4),a_(5)} , is equal to 44. Statement-2: The number of derangement for n objects is n! sum_(r=0)^(n)((-1)^(r))/(r!) .

Statement-1: If f:{a_(1),a_(2),a_(3),a_(4),a_(5)}to{a_(1),a_(2),a_(3),a_(4),a_(5)} , f is onto and f(x)nex for each x in{a_(1),a_(2),a_(3),a_(4),a_(5)} , is equal to 44. Statement-2: The number of derangement for n objects is n! sum_(r=0)^(n)((-1)^(r))/(r!) .

Statement-1: If f:{a_(1),a_(2),a_(3),a_(4),a_(5)}to{a_(1),a_(2),a_(3),a_(4),a_(5)} , f is onto and f(x)nex for each x in{a_(1),a_(2),a_(3),a_(4),a_(5)} , is equal to 44. Statement-2: The number of derangement for n objects is n! sum_(r=0)^(n)((-1)^(r))/(r!) .