Question

Find the sum of all 3 digit natural
numbers, which are divisible by 8.​

Answer 1

Answer :-

67,200

Explanation:-

The 3 digit number which is divisible by 8 are :-

104, 112 , 120 .............992.

The numbers are in A. P

The numbers are in A. Pso,

a = 104

a = 104 d = 112 - 104

a = 104 d = 112 - 104d = 8

a = 104 d = 112 - 104d = 8l = 992

First we have to find the number of terms,

$\huge \boxed{l = a+(n-1) d}$

→$992 = 104 + (n-1) 8$

→$992= 104 + 8n-8$

→$992 = 96 + 8n$

→$992 -96 = 8n$

→$896 = 8n$

→$n = \dfrac{896}{8}$

→$n = 112$

Now,

By using sum formula :-

$\huge \boxed{S_n = \dfrac{n}{2}[2a + l]}$

Put the given value,

→$S_{112} = \dfrac{112}{2}(2\times 104 + 992 )$

→$S_{112} = 56 ( 208 + 992 )$

→$S_{112} = 56 \times 1200$

→$S_{112} = 67,200$

hence, The sum of 3 digit number which are divisible by 8 is 67,200.