Though the problem looked difficult,it is basically a problem on A.P.Let no. of flowers bloomed be n_b and those fallen be n_f.Given,n_b-n_f=k(a constant)

No.of flowers that fell second time=5 means those that bloomed first time are 6.So,

k=6-0=6=n_b2-n_f2n_b2=11n_f3=10n_b3=16 and so on since it is given that flowers that fell this time is equal to those that bloomed previous time-1.

So,the no.of flowers bloomed are 6,11,16,21,.... and those that fell are 0,5,10,15,.....

Now,for first question,it is sum to 50 terms of first A.P.

for 2nd Q,it is sum to 25 terms of second A.P

for 3rd one,166=6+(n-1)5 gives the ans as 33.