`A_(1+(9))` stands for

If f(x)=prod_(n=1)^(100)(x-n)^(n(101-n)), where prod_(i=1)^(k)a_(i) stands for product a_(1)dot a_(2).........a_(k), then (f(101))/(f'(101))=(a)5050( b) (1)/(5050) (c) 10010 (d) (1)/(10010)

If a_(1),a_(2),...,.,a_(10) are positive numbers in arithmetic progression such that (1)/(a_(1)a_(2))+(1)/(a_(2)a_(3))+...*+(1)/(a_(9)a_(10))=(9)/(64) and (1)/(a_(1)a_(10))+(1)/(a_(2)a_(9))+......(1)/(a_(10)a_(1))=(1)/(10)((1)/(a_(1))+...+(1)/(a_(10))) then the value of (2)/(5)((a_(1))/(a_(10))+(a_(10))/(a_(1))) is

Consider an A.P. a_(1), a_(2), "……"a_(n), "……." and the G.P. b_(1),b_(2)"…..", b_(n),"….." such that a_(1) = b_(1)= 1, a_(9) = b_(9) and sum_(r=1)^(9) a_(r) = 369 , then

Find a_(1) and a_(9) if a_(n) is given by a_(n) = (n^(2))/(n+1)

Find a_(1) and a_(9) if a_(n) is given by a_(n) = (n^(2))/(n+1)

Given a_(1),a_(2),......,a_(12) are in G.P. If a_(4)*a_(9)=2019 then value of (a_(2)a_(11)+a_(3)a_(10)+a_(5)a_(8)+a_(6)a_(7))/(a_(4)a_(9)) is equal to

In an AP, as a_(7)=9 and a_(1)*a_(2)*a_(7) an to as is least,then the common difference is:

Find the respective terms for the following lowing Aps. (iii) a_(1) = 5, a_(4) = 9(1)/(2) , find a_(2), a_(3) .