Question

Find the sum of all natural numbers amongst first one thousand numbers which are neither divisible by 2 nor by 5.

Answer 1

let's do it

sum of n natural number is given by

$s(n) = \frac{n(n + 1)}{2}$

so we'll find sum of first 1000 numbers
which will result in

S(1000)=1000×1001/2 = 500,500 ....(1)

now, sum of terms divisible by 2 amongst 1st 1000 terms
s(divisible by 2)=500×501/2 = 125,250 ....(2)


now, sum of terms divisible by 5

s(divisible by 5) = 20×21/2 = 210 ....(3)



now your answer

s(divisible neither by 2 nor by 5)
= (1) - (2) - (3)
= 500500 - 125250 - 210
= 375040



hope that helped!
waiting for brainliest tag :P)

Answer 2

Answer:

20000

Step-by-step explanation:

1,2,3, ............, 999, 1000.The above list forms an AP with,first term, a = 1 and common difference, d = 1Now, number of terms, n = 1000We know that,Sn = n2[2a + (n−1)d] so,S1000 = 10002[2 + 999(1)] = 500 × 1001 = 500500.The list of numbers from 1 to 1000 divisible by 2 are :2,4, 6, 8, .........1000.The above list forms an AP with first term, a = 2 and common difference, d = 2Now, number of terms in the above AP, n = 500.Sn = n2[2a + (n−1)d] so,S500 = 5002[4 + 499(2)] = 250 × 1002 = 250500.The list of numbers from 1 to 1000 divisible by 5 are :5, 10, ........., 1000.The above list forms an AP with first term, a = 5 and common difference, d = 5Now, number of terms in the above AP, n = 200.Sn = n2[2a + (n−1)d] so,S200 = 2002[10 + 199(5)] = 100 × 1005 = 100500.The list of numbers from 1 to 1000 divisible by 10 are :10, 20, 30, ...........1000.The above list forms an AP with first term, a = 10 and common difference, d = 10Now, number of terms in the above AP, n = 100.Sn = n2[2a + (n−1)d] so,S100 = 1002[20 + 99(10)] = 50 × 1010 = 50500.sum of numbers from 1 to 1000 that are either divisible by 2 or by 5, S' =S500 + S200 − S100= 250500 + 100500 − 50500 = 300500sum of numbers from 1 to 1000 that are neither divisible by 2 nor by 5 =S1000 − S' = 500500 − 300500 = 200000