find sum of all natural number between 50 and 500 which are divisible by 5
Answers
Answer:
sum of all the natural numbers :
(a) between 100 and 1000 which are multiple of 5 = 98450
(b) between 50 and 500 which are divisible by 7 = 17696
(c) between 50 and 500 which are divisible by 3 and 5 = = 8325
Step-by-step explanation:
We need to find sum of all the natural numbers
a) between 100 and 1000 which are multiple of 5
Between 100 to 1000 means 101 to 999
105 , 110 , 115 , ...........................990 , 995
Sum = (105 + 110 + 115 +..................+ 990 + 995)
= 5 ( 21 + 22 + 23 +.......................+ 198 + 199)
21 to 199 are 179 number = n
a = 21 first number
d =1 common difference
l = 199 Last digit
sum= (n/2)(a+l)
=5 ( 179/2)(21 + 199)
= (895/2) 220
= 895 * 110
= 98450
b) between 50 and 500 which are divisible by 7
number 51 to 499
56 , 63 , ................... 490 , 497
Sum = 56 + 63 + ...........+ 490 + 497
= 7 ( 8 + 9 +.....................+ 70 + 71)
n = 64 a = 8 l = 71 d =1
= 7 (64/2)(8+71)
= 7 * 32 * 79
= 17696
(c) between 50 and 500 which are divisible by 3 and 5 =
divisible by 3 and 5 means divisible by 3 * 5 = 15 ( as no common factor)
60 , 75 , ........................., 480 , 495
Sum = (60 + 75 + ....................+ 480 + 495)
= 15 ( 4 + 5 +..........................+ 32 + 33)
n = 30 , a = 4 , l = 33
= 15 (30/2)(4 + 33)
= 15 * 15 * 37
= 8325
Step-by-step explanation:
Answer:3800
Step-by-step explanation:
Natural numbers from 51 to 100 will be in an arithmetic progression series with common difference of 1 and with first term of 51 and final term of 100
So let be 100 be the nth term so,
100 = 51 + (n-1) solving we will get the final count is 50
So now we can find out the sum according to A.P. rules
As sum = n/2 *( 2*a + (n-1)/2)
Putting n= 50 and a= 51 we will the required answer which will be 3800