Question

Question #09e08

Answer 1

Let n be the smallest of 6 consecutive number.So sum of 6 consecutive numbers forming an AP series of common difference 1 will be

`=6/2(2*n+(6-1)*1)=6n+15`

By the given condition

`6n+15=837`

`=>n=(837-15)/6=137`

So 6th number`=137+(6-1)*1=142`