Question

Find the sum of all integers between 200 and 500 which are divisible by 8

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

*** Excluding 200 & 500
(As per keyword in question, "between")

Smallest No. divisible by 8 & falling in the range = 208 = 26 × 8

Largest No. divisible by 8 & falling in the range = 496 = 62 × 8

•°• The series goes like:
208, 216, 224, ………, 480, 488, 496

For simplicity,
we take 8 common in all numbers, & then multiply later when sum is obtained.

Now, series will be
26, 27, 28, …………, 61, 62

Here, m = 26
& n = 62
•°• m-n+1= 37 ; m+n = 88

Substitute Values in 'S'
S = [88× 37]/2 = 1628

Now, Multiply with 8 to get required sum

•°• Required Sum = 1628 × 8 = 13024


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:)


Hope it helps