Question

Find the sum of all natural numbers with 300 and 500 which are divisible by 12

Answer 1

Hi there!

Here's the answer:

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First Note this formula

¶ Sum of Numbers 'S' from m to n is given by

S = [ (m+n)(n-m+1) ] /2

*** Here m<n.
Smaller/first No. = m
Larger/last No. = n

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¶¶¶ SOLUTION:

Given Range = 200 to 500

Smallest No. divisible by 12 & falling in the range = 204 = 17 × 12

Largest No. divisible by 12 & falling in the range = 492 = 41 × 12

•°• The series goes like:
204, 216, 228, ……… , 480, 492

For simplicity,
we take 12 common in all numbers, & then multiply later when sum is obtained.

Now, series will be
17, 18, 19, …………, 40, 41

Here, m = 17
& n = 41
•°• n-m+1= 25 ; m+n = 58

Substitute Values in 'S'

S = [58× 25]/2 = 1450/2 = 725

Now, Multiply with 12 to get required sum

•°• Required Sum = 725 × 12 = 8700

$\underline{\underline{Required\: Sum = 8700}}$

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:)

Hope it helps