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`

Options

A. 4:1
B. 6:1
C. 8:1
D. 10:1

Answer

B

Explanation

Let a be the first term and d be the common difference of the A.P. Then,
`S_(2n)=3S_(n)`
`rArr" "(2n)/(2){2a+(2n-1)d}=(3n)/(2){2a+(n-1)d}rArr2a=(n+1)d`.
`:." "(S_(3)n)/(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])=6`