Question

find the sum of all the 3 digit which are multiples of 8

Answer 1

192 is the multiple of 8

sum of 192=1+9+2=12

I hope answer is helpful!!

Answer 2

First three digit term divisible by 8 = 104

Last three digit term divisible by 8 = 992.

We know that tn = a + (n - 1) * d

= > 992 = 104 + (n - 1) * 8

= > 992 - 104 = 8n - 8

= > 888 + 8 = 8n

= > n = 112.

Now,

Sum of required series sn = (n/2)[a + l]

= > (112/2)[104 + 992]

= > (56[1096]

= > 61376.

Therefore, the sum of all three digit multiples of 8 = 61376.

Hope this helps!