Question

find the sum of all natural numbers between 100 and 1000 which are multiples of 3

Answer 1

Natural Numbers between 100 to 1000 which is multiples of 3 are ,
102,105,108,......999.

Here,

A = 102

D = 3

Tn = 999

a+(n-1)×d = 999

102+(n-1)×3 = 999

102+3n-3 = 999

3n = 999-102+3

3n =894.

n=894/3=298

Sn=N/2[2a+(n-1)×d]

SOLVE THIS YOU WILL GET YOUR ANSWER

Answer 2

The sum of all natural numbers between 100 and 1000 which are multiples of 3 is 1,65,150.

Given:

• All natural numbers between 100 and 1000.

To find:

• Find the sum of multiples of 3.

Solution:

Formula to be used:

1. nth term of A.P.$\bf a_n = a + (n - 1)d$
2. Sum of n terms of AP: $\bf S_n = \frac{n}{2} (a + l)$

Step 1:

Write the multiples of 3 between 100 and 1000.

102, 105, 108,...,999

Step 2:

Constitute AP.

Here,

First term of A.P. is a= 102

Last term is l= 999

Common difference = 3.

Step 3:

Find number of terms of AP.

$999 = 102 + (n - 1)3$

or

$3(n - 1) = 999 - 102$

or

$3(n - 1) =897$

or

$n - 1 = 299$

or

$\bf n = 300$

Step 4:

Find sum of all terms.

$S_{300} = \frac{300}{2} (102 + 999)$

or

$S_{300} = 150(1101)$

or

$\bf S_{300} = 165150$

Thus,

Sum of all natural numbers between 100 and 1000 which are multiples of 3 is 165150.

Learn more:

1) Find the sum of all 3 digit natural multiples of 6.

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2) Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 166833

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