Let `S_n` denote the sum of the first `n` tem of an A.P. If `S_(2n)=3S_n` then prove that `(S_(3n))/(S_n) =6.`

Explanation

Given, `S_(2n)=3S_n`

`⇒(2n)/2[2a+(2n−1)d]=(3n)/2[2a+(n−1)d]`

`⇒4a+(4n−2)d=6a+(3n−3)d`
or `2a=(n+1)d`

Now,
`(S_(3n))/(S_n)=((3n)/2[2a+(3n−1)d])/(n/2[2a+(n−1)d])`

`=(3[(n+1)d+(3n−1)d])/([(n+1)d+(n−1)d])`

`=(3[4nd])/([2nd])=6`