e. Find the Natural Number from 1 and 500 which are
divisble by 2 or 5
Answers
Answer:
Hi ,
1 ) 2 , 4 , 6 ,.... , 498 are the numbers
divisible by 2 between 1 to 500
They are in A.P
first term = a = 2
Common difference = d = 4 - 2 = 2
last term = l = 498
Let number of terms = n
a + ( n - 1 )d = l
2 + ( n - 1 ) 2 = 498
( n - 1 ) 2 = 496
n - 1 = 248
n = 249
Therefore ,
Sum of the terms = n/2 [ a + l ]
= 249 /2 [ 2 + 298 ]
= 249 /2 × 300
= 74700/2
= 37350
2 ) 5 , 10 , 15 , .... , 495 are the
numbers divisible by 5 between 1 and
500
They are in A.P
a = 5 , d = 5
l = 495
a + ( n - 1 )d = 495
5 + ( n - 1 ) 5 = 495
( n - 1 )5 = 490
n - 1 = 490/5
n - 1 = 98
n = 99
Therefore b,
Sum = n/2 [ a +l ]
= 99/2 [ 5 + 495 ]
= 99/2 × 500
= 50500/2
= 25250
3 ) 10 , 20 , 30 , ....490 are
numbers which are divisible by 2 and 5 .
There are in A.P
a = 10
d = 20 - 10 = 10
l = 490
let number of terms = n
a + ( n - 1 )d = l
10 + ( n - 1 ) 10 = 490
( n - 1 ) 10 = 480
n - 1 = 480/10
n - 1 = 48
n = 49
sum = n/2 [ a + l ]
= 49 /2 [ 10 + 490 ]
= 49/2 × 500
= 24500/2
= 12250
I hope this helps you.
:)
mark brainliest if helpful