Question

The unitdigit of the number (1)33+2(33)+3(33)+4(33)+........89(33) is

Answer 1

Answer:

taking 33 as common

we get

33 (1+2+34.....89).

in bracket find the sum of 89 natural numbers

as n(n+1)/2

33(89(90)/2

33 x 89 x 45

132165

thus the unit digit is 5

Answer 2

Answer:

a=1

d=a2-a1=2-1=1

an=89

an=a+(n-1)d

89=1+(n-1)1

89-1/1=n-1

88/1=n-1

88+1=n

n=89