Let `S_n` denote the sum of first `n` terms of an A.P. If `S_(2n)=3S_n ,` then find the ratio `S_(3n)//S_ndot`

Answer

6

Explanation

Given, `S_(2n)=3S_(n)`
`rArr(2n)/2[2a+(2n-1)d]=3n/2[2a+(n-1)d]`
`rArr4a+(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`