Find the sum of all natural number from 1 to100 which are divisible by 4 or 5
Answers
Answer:
ANSWER
The integers from 1 to 100 which are divisible by 2
are 2,4,6,/...100 this forms an A.P with both first term and common difference =2
⇒ 100=2+(n−1)2 ⇒n=50
∴ 2+4+6+....+100=
2
50
[2(2)+(50−1)2]
=25(102)=2550
The integers form 1 to 100, which are divisible by 5 are 5,10,15...100 whose a=5, d=5
∴ 100=5+(n−1) ⇒ 5n=100 or n=20
∴ 5+10+...+100=
2
20
[2(5)+(20−1)5]
=10(10+95)=1050
The integers which are divisible by 2 and 5 are 10,20,30...100 terms an A.P with a=10,d=10.
∴ 100=10+(n−1)10 ⇒ 100−10=(n−1)
⇒ 90=10(n−1)
⇒ n=10
∴ 10+20+....+100=
2
10
[2(10)+(10−1)10]
=5[20+90]=550
∴ Required sum =2550+1050−550
=3050
Thus, sum of the integers from 1 to 100, which are divisible by 4 or 5 is 3050.
Step-by-step explanation:
hey mate here is your answer
Answer:
1150
Step-by-step explanation:
1st AP
4, 8, 12, .......... , 100
a = 4, d = 4, an = 100
an = a + (n - 1)d
100 = 4 + (n - 1)4
96 = (n - 1)4
24 = n - 1
n = 25
Sn = n/2 (a + l)
Sn = 25/2 (4 + 100)
Sn = 25/2 × 104
Sn = 25 × 52
Sn = 1300
2nd AP
5, 10, 15, ........., 100
a = 5, d = 5, an = 100
an = a + (n - 1)d
100 = 5 + (n - 1)5
95 = (n - 1)5
19 = n - 1
n = 20
Sn = n/2 (a + l)
Sn = 20/2 (5 + 100)
Sn = 10 × 105
Sn = 1050
3rd AP
20, 40, 60, ........ , 100
a = 20, d = 20, an = 100
an = a + (n - 1)d
100 = 20 + (n - 1)20
80 = (n - 1)20
4 = n - 1
n = 5
Sn = n/2 (a + l)
Sn = 5/2 (20 + 100)
Sn = 5/2 × 120
Sn = 5 × 60
Sn = 1200
Required answer = 1300 + 1050 - 1200
= 1150