Question

See the sum of all three digits that are multiples of six See the sum of all three digits that are multiples of six​

Answer 1

Answer:

$a_{n}=a+(n-1)d$  --- (1)

$S_{n} =\frac{n}{2} [2a+(n-1)d]$  --- (2)

Explanation:

The question is to find the sum of an AP.

First three digit divisible by 6, hence a=102.

common difference=6, The last digit of AP is 996.

Hence solving (1) , we get

n=150

Then solving (2) , we get

Sum= 82,350

Hope it helped...