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

Given, `S_(n)="Sum of first n terms of an AP."`
`S_(n)=(h)/(2)[2a+(n-1)d]" or "S_(2n)=(2n)/(2)[2a+(2n-1)d]`
Similarly, `S_(3n)=(3n)/(2)[3a+(3n-1)d]`
According to direction, `3S_(n)=2S_(2n)`
Putting the value of `S_(n)` and `S_(2n)` in above equation.
`3((n)/(2))[2a+(n-1)d]=2((n)/(2))[2a+(2n-1)d]`
`6a+3(n-1)d=4a+2(n-1)d`
`2a=d(n+1)`
`:.S_(n)=(n)/(2)[d(n+1)+d(n-1)]`
`=(n)/(2)[dn+d+dn-d]`
`=(n)/(2)(2"dn")=n^(2)d`
Now, `S_(2n)=n[d(n+1)+(2n-1)d]=3n^(2)d`
`S_(3n)=(3n)/(2)[d(n+1+3n-1)]=6n^(2)d`
Hence, `(S_(3n))/(S_(n))=(6n^(2)d)/(n^(2)d)=(6)/(1)=6:1`