Find the sum of all three digit natural numbers which are multiples of 4.
Answers
Answered by
1
Answer:
I don't know I don't know duniya tere te merdi dudn waste :) this song is for maths +_+
Answered by
0
Answer:
44550
First three digit number divisible by 11 is 110
Last three digit numbers divisible by 11 is 990
∴a=110 a
n
=990 d=11 x=2
∴ No. of =2
Terms
a
n
=a+(n−1)d
990=110+(n−1).11
11
880
=x−1
80=x−1
x=81
∴ sum of all terms of A.P is given by
S
n
=[a+a
n
]×
2
n
=[110+990]×
2
81
=1100×
2
81
=550×81
=44550
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