The sum of all natural numbers from 100 to 300 excluding those, which are divisible by 4, is
Answers
Step-by-step explanation:
Method to Solve :
sum of numbers divisible by 4
= 100 + 104 + 108 + ... + 300
total numbers = 51
Sn = n/2[2a + (n – 1)d]
= 51/2[200 + 50 � 4]
= 51/2[400]
= 51 � 200
= 10200
sum of numbers divisible by 5
= 100 + 105 + 110 + 115 + ... + 300
total numbers = 41
Sn = n/2[2a + (n – 1)d]
= 41/2[200 + 40 � 5]
= 41/2[400]
= 41 � 200
= 8200
but above additions contain numbers which are divisible by 20 (LCM of 4 and 5)
to be subtracted.
sum of numbers divisible by 20
= 100 + 120, + 140, + ... + 300
total numbers = 11
= 11/2[200 + 10 � 20]
= 11 � 200
= 2200
required sum = 10200 + 8200 – 2200
= 16200
I hope it's help you
Answer:
19800
Step-by-step explanation:
100,104,108,……..300 it is in Arithmetic progression
first term,a=100
Common difference,d=4
Last term=300
Now, we know the nth term in the A. P can be written as a+(n-1)d=300
n-1=(300–100)/4=50
Therefore no of digits, n=51
Sum of A. P series will be
= a+(a+d)+(a+2d)+(a+3d)+………..(a+(n-1)d)
=(n*a)+d[0+1+2+…….+(n-1)]=(na)+d[(n-1)n/2]
=(51*100)+4[50*51/2]
=5100+5100
=20200
Sum of all natural numbers between 100 and 300(divisible by 4) will be =20200–100–300=19800.