what is the sum of the no.s that lie b/w 602 , 901 which are not dvisible by four
Answers
Answer:
Hi
Step-by-step explanation:
First we find sum of all natural numbers which are divisible by 4
604, 608,............900
This is an A.P with a=604 and d=4
Number of terms,
n=\frac{l-a}{d}+1n=
d
l−a
+1
n=\frac{900-604}{4}+1n=
4
900−604
+1
n=\frac{296}{4}+1n=
4
296
+1
n=74+1n=74+1
n=75n=75
Sum of natural numbers between 602 and 902 which are divisible by 4 is
S_n=\frac{n}{2}[a+l]S
n
=
2
n
[a+l]
S_{75}=\frac{75}{2}[602+902]S
75
=
2
75
[602+902]
S_{75}=\frac{75}{2}[1504]S
75
=
2
75
[1504]
S_{75}=75*752S
75
=75∗752
S_{75}=56400S
75
=56400
Now we find sum of all natural numbers from 603 to 901
=603+603+........+901=603+603+........+901
=(1+2+......+901)-(1+2+.....+602)=(1+2+......+901)−(1+2+.....+602)
=\frac{901*902}{2}-\frac{602*603}{2}=
2
901∗902
−
2
602∗603
=(901*451)-(301*603)=(901∗451)−(301∗603)
=(901*451)-(301*603)=(901∗451)−(301∗603)
=406351-181503=406351−181503
=224848=224848
The required sum
=(sum of all natural numbers from 603 to 901)-(sum of all natural numbers between 602 and 902 which are divisible by 4)
=224848-56400
=168448